By a well-known result if Zsigmondy, with a few exceptions, for $k>1$, there is a prime $r$ that divides $p^k-1$, but does not divide $p^i-1$ for any $i<k$. Then a Singer cyclic group in ${\rm PGL}(k,q)$ is contained in the centralizer $C$ of a Sylow $r$-subgroup $R$ of ${\rm PGL}(k,q)$.
Since the inverse image of $R$ in ${\rm GL}(k,q)$ is a direct product $\hat{R} \times Z$, where $Z$ is the group of scalar matrices, which has order $q-1$, coprime to $r$, and $\hat{R} \in {\rm Syl}_r({\rm GL}(k,q))$, the inverse image in ${\rm GL}(k,q)$ of $C$ centralizes $\hat{R}$.
Now $\hat{R}$ acts irreducibly on ${\mathbb F}_q^k$, so by Schur's lemma, its centralizer in ${\rm GL}(k,q)$ is the multiplicative group of a division algebra over ${\mathbb F}_q$, which is commutative since the field is finite. Since this centralizer contains a Singer cyclic of order $q^k-1$, this division algebra must be ${\mathbb F}_{q^k}$, and so this Singer cycle is the complete centralizer of $\hat{R}$ in ${\rm GL}(k,q)$.
So its image in ${\rm PGL}(k,q)$, which is original Singer cycle, must be the full centralizer $C$ of $R$, and now it follows from Sylow's Theorem that any two such Singer cycles are conjugate in ${\rm PGL}(k,q)$.
The exceptional cases in which the Zsigmondy prime does not exist are when $q=2$, $k=6$, in which case $Z=1$ so there is nothing to prove, and when $k=2$ and $q+1$ is a power of $2$, but in that case a Sylow $2$-subgroup of ${\rm PGL}(k,q)$ is dihedral of order $2(q+1)$, and the Singer cycle is its unique maximal cyclic subgroup, so the result follows again from Sylow's Theorem.
${\bf Added\ later}$: The above argument is for ${\rm PGL}(k,q)$, not ${\rm P \Gamma L}(k,q)$. I believe that the result is valid in ${\rm P \Gamma L}(k,q)$, and I will just sketch an argument.
So let $q=p^e$ with $p$ prime. We will choose our Zsigmondy prime $r$ such that $r$ divides $q^k-1 = p^{ke}-1$, but $r$ does not divide $p^m-1$ for any $m < ke$ (and note that this means we have a couple more exceptional cases, $q^k = 4^3,8^2$ that have to be handled separately, possibly by computer).
Note that $r$ cannot divide $e$, since if $e=rs$ then $r$ divides $p^{krs} - p^{ks}$ and hence $r$ divides $p^{ks}-1$, contrary to assumption. So our Sylow $r$-subgroup $R$ of $G$ still lies in ${\rm PGL}(k,q)$.
The group ${\rm \Gamma L}(k,q)$ is defined as the semidirect product ${\rm GL}(k,q) \rtimes \langle \sigma \rangle$, where $\sigma$ induces the field automorphism of order $e$ that acts by applying $x \mapsto x^e$ to matrix entries. This group projects onto ${\rm P\Gamma L}(k,q)$.
Now ${\rm \Gamma L}(k,q)$ embeds into $ {\rm GL}(ke,p)$. By the same argument as before, the inverse image in ${\rm GL}(ke,p)$ of the centralizer $C$ of $R$ in ${\rm PGL}(ke,p)$ is cyclic of order $q^k-1$, which is the same as its centralizer in ${\rm GL}(k,q)$, and hence also the same as its centralizer in ${\rm \Gamma L}(k,q)$.
So $C$ is the centralizer of $R$ in ${\rm P \Gamma L}(k,q)$, and the argument proceeds as before.
I know this proof is more complicated than you might prefer, and it is possible that there is an easier proof. But I have some previous experience with proving results in these semilinear groups, and the use of Zsigmondy primes is a standard technique.
