For simplicity, we will start with the case $z = 1$.
We will assume $k > 0$ and let $\cot\theta = k$ and $\mu_{\pm} = k \pm \sqrt{k^2+1}$.
The CF at hand has the form $$
\def\CF{\mathop{\LARGE\mathrm K}}
\CF_{\ell=1}^{\infty} \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_\ell}
\quad\text{ where }\quad
\gamma_0 = 1\quad\text { and }\quad
\begin{cases}
\alpha_\ell &= \ell m + 1\\
\beta_\ell &= k m (2\ell - 1)\\
\gamma_\ell &= \ell m - (m+1)
\end{cases}
\quad\text{ for }\quad\ell > 0
$$
Consider the associated linear recurrence relation:
$$\alpha_\ell f_{\ell-1} - \beta_\ell f_\ell = \gamma_\ell f_{\ell + 1}\quad\text{ for }\ell > 0\tag{*1}$$
Let $(u_\ell)_{\ell=0}^\infty$ and $(v_\ell)_{\ell=0}^\infty$ be the two solutions satisfying $(u_0,u_1) = (1,0)$ and $(v_0,v_1) = (0,1)$.
In terms of $u$ and $v$, it is easy to verify the convergents of CF
satisfy
$$L_n \stackrel{def}{=} \CF_{\ell=1}^n \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_{\ell}} = -\frac{u_{n+1}}{v_{n+1}}
\quad\implies\quad
L_\infty \stackrel{def}{=} \CF_{\ell=1}^\infty \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_{\ell}} = -\lim_{n\to\infty} \frac{u_n}{v_n}$$
Notice $\alpha_\ell,\gamma_\ell \ne 0$ for all $\ell$ and following limits exist
$$
\alpha \stackrel{def}{=}
\lim\limits_{\ell\to\infty}\frac{\alpha_\ell}{\ell m} = 1,
\quad
\beta \stackrel{def}{=}
\lim\limits_{\ell\to\infty}\frac{\beta_\ell}{\ell m}
= 2k
\quad\text{ and }\quad
\gamma \stackrel{def}{=}
\lim\limits_{\ell\to\infty}\frac{\gamma_\ell}{\ell m}
= 1
$$
Furthermore, the associated polynomial
$$\alpha - \beta\lambda - \gamma\lambda^2 = 0 \quad\iff\quad
\lambda^2 +2k\lambda - 1 = 0$$
have roots $\lambda_{\pm} = -k \pm \sqrt{k^2+1} = \frac{1}{\mu_{\pm}}$ whose modulus are distinct.
By Poincare-Perron theorem${}^{\color{blue}{[1]}}$, $(*1)$ has a pair of fundamental solutions $(p^{+}_\ell)_{\ell=0}^\infty$ and $(p^{-}_\ell)_{\ell=0}^\infty$ satisfying
$\lim\limits_{n\to\infty} \frac{p^{\pm}_{\ell+1}}{p^{\pm}_{\ell}} = \lambda_{\pm}$.
Expanding $u$, $v$ in terms of $p^{\pm}$ and notice $|\lambda_{-}| > 1 > |\lambda_{+}|$, we have
$$\begin{cases}
u_\ell &= A p^{+}_\ell + B p^{-}_\ell\\
v_\ell &= C p^{+}_\ell + D p^{-}_\ell
\end{cases}
\quad\implies\quad
L_\infty = -\lim_{n\to\infty} \frac{A p^{+}_\ell + B p^{-}_\ell}{C p^{+}_\ell + D p^{-}_\ell} = -\frac{B}{D}$$
Construct a solution $(w_\ell)_{\ell=0}^\infty$ of $(*1)$ by setting $w_\ell = u_\ell + L_\infty v_\ell$, we have
$$w_\ell = ( A p^{+}_\ell + B p^{-}_\ell ) - \frac{B}{D}( C p^{+}_\ell + D p^{-}_\ell ) = \frac{AD-BC}{D} p^{+}_\ell
\quad\implies\quad
\lim_{\ell\to\infty} \frac{w_{\ell+1}}{w_\ell} = \lambda_{+}
$$
The corresponding OGF $\displaystyle\;\quad w(t) \stackrel{def}{=} \sum_{\ell=0}^\infty w_\ell t^\ell\quad$ will be analytic for $t < |\mu_{+}|$.
Let $a = \frac{1}{m}$, one can rescale $(*1)$ to the form:
$$(\ell + a) w_{\ell-1} - k(2\ell-1) w_{\ell} + (a + 1 - \ell) w_{\ell+1} = 0\quad\text{ for }\quad \ell > 0\tag{*2}$$
Let us first treat $a$ as a complex parameter with $-\frac32 < \Re a < -1$.
Multiply the $\ell^{th}$ entry of $(*2)$ by $t^\ell$ and start to sum from $\ell = 1$. If we let $g(t) = w(t) - 1$, we find
$$\begin{align}
& \left[t\frac{d}{dt} + a\right]\big\{t(1+g(t))\big\}
-k\left[2t\frac{d}{dt} - 1\right]g(t)
+ \left[a+1 - t\frac{d}{dt}\right]\left\{\frac{g(t)}{t} - L_{\infty}\right\}
= 0\\
\iff &
(t^2 - 2kt - 1)g'(t)
+ \left((a+1)t + k + \frac{a+2}{t}\right) g(t)
+ (a+1)(t - L_\infty) = 0\\
\iff & g'(t) + \left[ -\frac{a+2}{t} + \frac{a+\frac32}{t - \mu_{+}} + \frac{a + \frac32}{t - \mu_{-}}\right] g(t) = (a+1)\frac{t - L_\infty}{1+2kt - t^2}\\
\iff &
\frac{d}{dt}\left[\frac{(1+2kt-t^2)^{a+\frac32}}{t^{a+2}} g(t)\right]
= (a+1)\frac{(1+2kt-t^2)^{a+\frac12}}{t^{a+2}}(t-L_\infty)\\
\implies &
w(t) = 1 + (a+1)\frac{t^{a+2}}{(1+2kt-t^2)^{a+\frac32}}\left[\int_0^t
\frac{(1+2ks-s^2)^{a+\frac12}}{s^{a+2}}(s-L_\infty) ds\right]
\end{align}
$$
For the range of choice of $a$, the coefficient in front of the square bracket is singular at $t = \mu_{-}$. Since $|\mu_{-}| < |\mu_{+1}|$, $w(t)$ is regular there. This means the integral in the square bracket need to vanish at $t = \mu_{-}$. From this, we can deduce
$$L_{\infty} = \frac{\Lambda(a,-a)}{\Lambda(a,-a-1)}
\quad\text{ where }\quad
\Lambda(a,b) \stackrel{def}{=}
\int_0^{\mu_{-}}
s^{b-1}(1+2ks-s^2)^{a+\frac12} ds
$$
Let $c = a + b + \frac32$ and $\eta = \frac{\mu_{-}}{\mu_{+}}$, the integral $\Lambda(a,b)$ can be evaluated as
$$\begin{align}\Lambda(a,b)
& =
\int_0^{\mu_{-}}
s^{b-1}\left(1 - \frac{s}{\mu_{+}}\right)^{a+\frac12}\left(1 - \frac{s}{\mu_{-}}\right)^{a+\frac12} ds\\
&= \mu_{-}^b \int_0^1 s^{b-1}(1-s)^{c-b-1}\left(1-\eta s\right)^{a+\frac12} ds\\
&=^{\color{blue}{[2]}} \mu_{-}^b \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}{}_2F_1\left(-a-\frac12,b;c; \eta\right)\\
&=^{\color{blue}{[3]}} \mu_{-}^b(1 - \eta)^{-b} \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}
{}_2F_1\left(c+a+\frac12,b;c; \frac{\eta}{\eta - 1}\right)\\
\end{align}
$$
Substitute this back into expression of $L_\infty$ and simplify, we get
$$L_\infty = -\frac{2(a+1)\mu_{-}}{1 - \eta}
\frac{
{}_2F_1\left(a+2,-a;\frac32; \frac{\eta}{\eta-1}\right)
}{
{}_2F_1\left(a+1,-(a+1);\frac12; \frac{\eta}{\eta-1}\right)
}
$$
Notice
$$
\frac{\mu_{-}}{1-\eta} = -\frac{1}{2\sqrt{k^2+1}} = -\frac{\sin\theta}{2}
\quad\text{ and }\quad
\frac{\eta}{\eta - 1} = \frac{\sqrt{k^2+1}-k}{2\sqrt{k^2+1}} = \frac{1-\cos\theta}{2}
$$
We get
$$L_\infty =
\frac{(a+1)\sin\theta
{}_2F_1\left(a+2,-a;\frac32; \frac{1-\cos\theta}{2}\right)
}{
{}_2F_1\left(a+1,-a-1,\frac12; \frac{1-\cos\theta}{2}\right)
}
$$
Compare this with following representation of
generalized Chebyshev's polynomial in terms of hypergeometric functions.
$$\begin{cases}
T_a(\cos x) &= {}_2F_1\left(a,-a; \frac12; \frac12(1-\cos x)\right) = \cos(a x)\\
U_a(\cos x) &= (a+1){}_2F_1\left(a+2,-a; \frac32; \frac12(1 - \cos x)\right) = \frac{\sin((a+1)x)}{\sin x}
\end{cases}
$$
We find
$$L_\infty = \tan((a+1)\theta) = \tan\left((a+1)\tan^{-1}\frac{1}{k}\right)$$
Up to what I see, there is nothing to stop us from analytic continuing this result
back to $a \in (0,1)$. This implies
$$\Theta(k,1,m) = \tan\left(\frac{m+1}{m}\tan^{-1}\frac1k\right)$$
Back to the case $z \ne 1$. The CF can be obtained by scaling $\alpha_\ell, \gamma_\ell$ by $\sqrt{z}$ for all $\ell > 0$ and $\gamma_0$ by $\frac{1}{\sqrt{z}}$. Since $(*1)$ is invariant when one rescale $\alpha_\ell, \beta_\ell, \gamma_\ell$ by same factor, scaling $\alpha_\ell, \gamma_\ell$ by $\sqrt{z}$ is equivalent to replace the $k$ in $\beta_\ell$ by $\frac{k}{\sqrt{z}}$.
As long as $z \in \mathbb{C} \setminus (-\infty, 0 ]$, we have $\Re \sqrt{z} > 0$. One can verify the corresponding $\lambda_{\pm}$ have distinct modulus and above arguments remain valid.
As a result, the CF at hand has following closed form expression:
For $k > 0$, $m > 1$ and $z \in \mathbb{C} \setminus (-\infty,0]$, $$\Theta(k,z,m) =
\frac{1}{\sqrt{z}}\Theta\left(\frac{k}{\sqrt{z}},1,m\right) =
\frac{1}{\sqrt{z}}\tan\left[\frac{m+1}{m}\tan^{-1}\left(\frac{\sqrt{z}}{k}\right)\right]$$
The rest is obvious.
Notes
$\color{blue}{[1]}$ -
For more details of Poincare-Perron theorem and a proof of it, please consult Chapter 8 of Saber Elaydi's book An Introduction to Difference Equations.
$\color{blue}{[2]}$ - For $|t| < 1$ and $\Re c > \Re b > 0$, we have following
integral formula by Euler
$$\int_0^1 x^{b-1}(1-x)^{c-b-1}(1-tx)^{-a} dx = \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)} {}_2F_1(a,b;c;t)$$
$\color{blue}{[3]}$ - We are using following Pfaff transformation here.
$${}_2F_1(a,b;c;t) = (1-t)^{-b} {}_2F_1\left(c-a,b;c;\frac{t}{t-1}\right)$$
