Preliminary note.
You made a typo in the first equation : the coefficient $\gamma$ should be replaced by $1/\gamma$, according to the standard parametrization of Milburn's master equation (and the subsequent solution you provided).
First remarks.
It is to be noted that the right-hand side of the equation of motion is time-independent and linear with respect to the density matrix. In consequence, this equation can be recasted as $\dot{\rho} = X\rho$, where $X$ is a linear operator $-$ actually a superoperator, since it acts on the space of operators to which $\rho$ belongs.
This form is practical, since it allows to solve the equation in the very same way as the Schrödinger equation with the time evolution operator, i.e. $\partial_t|\psi(t)\rangle = -\frac{i}{\hbar}\hat{H}|\psi(t)\rangle \Rightarrow |\psi(t)\rangle = \hat{U}(t)|\psi(0)\rangle$, where $\hat{U}(t) = e^{-it\hat{H}/\hbar}$.
In the present case, one gets $\rho(t) = e^{tX}\rho(0)$. On a side note, $X$ is the generator of the flow $e^{tX}$, even if the word propagator is usually preferred for the latter in the physics litterature. But how to explicit this operator $X$ in the present situation ? In order to do so, we need to make a detour by Lie theory, whose ultimately purpose deals with the relations between flows and generators.
Recalls on Lie theory.
Each Lie group $G$ is associated to a Lie algebra $\mathfrak{g}$; the exponential map and the derivative make the link between them $-$ be aware that they are not bijections, but surjections only. Indeed, $\forall g \in G$ $\exists X \in \mathfrak{g}$ s.t. $g = e^{tX}$ and, vice versa, $\forall X \in \mathfrak{g}$ $\exists g \in G$ s.t. $X = \left.\frac{\mathrm{d}g}{\mathrm{d}t}\right|_{t=0}$.
These two spaces act on themselves, as follows :
$$
\left\{
\begin{array}{r}
\mathrm{Ad} : G \times G \rightarrow G, (g,h) \mapsto \mathrm{Ad}_gh := ghg^{-1} \\
\mathrm{ad} : \mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g}, (X,Y) \mapsto \mathrm{ad}_XY := [X,Y]
\end{array}
\right.
$$
These actions are called the adjoint actions, from which you will have recognized conjugation and commutator respectively. They are themselves related by the exponential map mentioned supra, so that $\mathrm{Ad}_{\exp(A)} = \exp(\mathrm{ad}_A)$; this relation is nothing else than the more common following formula in physics :
$$
e^ABe^{-A} = B + [A,B] + \frac{1}{2!}[A,[A,B]] + \ldots
$$
Resolution of Milburn's equation.
We understand from what is discussed above that the equation of motion can be reformulated as follows :
$$
\dot{\rho} = -i[H,\rho] - \frac{1}{2\gamma}[H,[H,\rho]] = -i\,\mathrm{ad}_H\rho - \frac{1}{2\gamma}\mathrm{ad}_H^2\rho,
$$
hence $X = -i\,\mathrm{ad}_H - \frac{1}{2\gamma}\mathrm{ad}_H^2$ and thus $\rho(t) = e^{tX}\rho(0)$. Now, it is easy to show that $\mathrm{ad}_H$ and $\mathrm{ad}_H^2$ commute, so that the exponential of the propagator can be split into two parts, i.e. $e^{tX} = \exp(-it\,\mathrm{ad}_H) \exp\left(-\frac{t}{2\gamma}\mathrm{ad}_H^2\right)$.
Let's work on the operator $\mathrm{ad}_H^2$. One has :
$$
\mathrm{ad}_H^2B = [H,[H,B]] = [H,HB-BH] = H^2B - 2HBH + BH^2 = (L_H - 2C_H + R_H)B,
$$
where $L_HB := H^2B$, $C_HB := HBH$ and $R_HB := BH^2$. It can be checked straightforwardly that those three opeartors commute pairwise, that is why the last exponential can also be split into three separate parts, i.e.
$$
\exp\left(-\frac{t}{2\gamma}\mathrm{ad}_H^2\right) = \exp\left(-\frac{t}{2\gamma}L_H\right) \exp\left(\frac{t}{\gamma}C_H\right) \exp\left(-\frac{t}{2\gamma}R_H\right),
$$
hence in the end
$$
\rho(t) = e^{-it\,\mathrm{ad}_H} e^{-\frac{t}{2\gamma}L_H} e^{\frac{t}{\gamma}C_H} e^{-\frac{t}{2\gamma}R_H} \rho(0).
$$
The last three exponentials can be now applied to the initial condition $\rho(0)$ with the help of their series expansions :
$$
\begin{array}{rcl}
e^{-\frac{t}{2\gamma}L_H} e^{\frac{t}{\gamma}C_H} e^{-\frac{t}{2\gamma}R_H} \rho(0)
&=& \displaystyle
\sum_{n\ge0} \frac{(-t/2\gamma)^n}{n!} L_H^n
\sum_{n\ge0} \frac{(t/\gamma)^k}{k!} C_H^k
\sum_{n\ge0} \frac{(-t/2\gamma)^m}{m!} R_H^m
\rho(0) \\
&=& \displaystyle
\sum_{n,k,m\ge0}
\frac{(-t/2\gamma)^n}{n!}
\frac{(t/\gamma)^k}{k!}
\frac{(-t/2\gamma)^m}{m!}
H^{2n} H^k\rho(0)H^k H^{2m} \\
&=& \displaystyle
\sum_{k\ge0}
\frac{(t/\gamma)^k}{k!}
e^{-\frac{t}{2\gamma}H^2}
H^k\rho(0)H^k
e^{-\frac{t}{2\gamma}H^2}
\end{array}
$$
Finally, the first exponential of the propagator is applied thanks to the identity mentioned above, namely $e^{-it\,\mathrm{ad}_H} = \mathrm{Ad}_{e^{-itH}}$, hence the final result you wanted to derive :
$$
\rho(t) = \sum_{k\ge0}
\frac{(t/\gamma)^k}{k!}
e^{-itH}
e^{-\frac{t}{2\gamma}H^2}
H^k\rho(0)H^k
e^{-\frac{t}{2\gamma}H^2}
e^{+itH},
$$
where you can identify $M_k := e^{-itH} e^{-\frac{t}{2\gamma}H^2} H^k$.
Final note.
It is to be noted that the Milburn's original model is Poissonian, so that the equation of motion for $\rho(t)$ is given by
$$
\dot{\rho} = \gamma\left(e^{-iH/\gamma} \rho e^{+iH/\gamma} - \rho\right)
= \gamma\left(e^{-\frac{i}{\gamma}\mathrm{ad}_H}-1\right)\rho,
$$
which can be solved explicitly with the same method, which leads to
$$
\rho(t) = \exp\left(\gamma t\left(e^{-\frac{i}{\gamma}\mathrm{ad}_H} - 1\right)\right)\rho(0)
$$
Actually, the master equation in your question corresponds to a $\mathcal{O}\left(\frac{1}{\gamma^2}\right)$-expansion of this model.
