How do I show that $\text{End}_R\mathbb{Z}^n\cong\mathbb{Z}$ for $R=\text{Mat}_n(\mathbb{Z})$? Let $R=\text{Mat}_n(\mathbb{Z})$ and $M=\mathbb{Z}^n$ the (left) $R$-module with action the matrix multiplication. 

How do I prove that $\text{End}_RM\cong\mathbb{Z}$?

Should I find an explicit isomorphism?

To rephrase your question, the elements of $End(M_R)$ are those elements of $R$ which commute with all other elements of $R$.
So, you are just looking for the center of a matrix ring.
Here are breadcrumbs to follow to get to this idea:
For any $S$-module $M$, we have $S\subseteq End(M_\Bbb Z)$ in a natural way (multiplication by elements of $S$ make additive maps.)
For any ring $S$, $End(S^n_S)\cong Mat_n(S)$.
Finally, $End(M_S)$ is, by definition, the subring of $End(M_\Bbb Z)$ whose elements all commute with the elements of $S$.
By slotting these with the specific situation you were given, you arrive at my original suggestion.
