Please, check my version:
As $z \in \mathbb{Z} \Rightarrow z = a + ib$, where $a, b \in \mathbb{R}$
Thus
$z(1 + \sqrt{-5}) = (a + ib)(1 + i \sqrt{5}) = a + ai \sqrt{5} + ib + {i}^{2} \cdot b \sqrt{5} = (a - b \sqrt{5}) + i(a \sqrt{5} + b) = 7 + i \sqrt{5}$
Solve the following system of equations:
\begin{cases} a - b \sqrt{5} = 7 \\ a \sqrt{5} + b = \sqrt{5} \end{cases} $\Rightarrow$ \begin{cases} a = b \sqrt{5} + 7 \\ (b \sqrt{5} + 7) \sqrt{5} + b = \sqrt{5} \end{cases} $\Rightarrow$ \begin{cases} a = 2 \\ b= -\sqrt{5} \end{cases} Thus, we get there exist the value $z \in \mathbb{Z}(z = 2 - i \sqrt{5})$ such that $7 + \sqrt{-5} = z + z \sqrt{-5}$
Thus, $7 + \sqrt{-5}$ is reducible in $\mathbb{Z} + \mathbb{Z} \sqrt{-5}$.