We can also rule out the case of commutative $R$ by appealing to the Artin–Procesi theorem: an Azumaya algebra of constant rank $(n+1)^2$ (e.g. $M_{n+1}(R)$) satisfies all the $\mathbb Z$-multilinear identities of $M_{n+1}(\mathbb Z)$ but no nonzero homomorphic image of it satisfies all the $\mathbb Z$-multilinear identities of $M_n(\mathbb Z)$.
It's perhaps worth noting that if R is a field, then there's a fairly straightforward way of proving that there is no injective ring homomorphism M_{n+1}(R) \to M_n(R). In fact, suppose we have a nonzero ring homomorphism M_{n'}(R) \to M_n(R). Then this allows us to view R^n as a left M_{n'}(R)-module. Now if R is a field, then M_{n'}(R) is simple, and so R^n decomposes into a finite direct sum of irreducible M_{n'}(R)-modules. It's a standard fact (and one that is easy to prove) that each such module is isomorphic to R^{n'}. We thus obtain an isomorphism R^n = R^{n'} \oplus \cdots \oplus R^{n'} of M_{n'}(R)-modules, and hence of R-vector spaces by restricting the action to the subring of scalar matrices. But then linear algebra allows us to conclude that n'|n. Nevermind. :)
Update: It's possible to have a nontrivial ring map $M_{n+1}(R) \to M_n(R)$ with $R$ finitely generated (and necessarily noncommutative). The idea, inspired by my previous mishap and wccanard's comment, is to find a finitely generated ring $R$ for which there is an isomorphism $R^{n+1} \cong R^n$ of left $R$-modules. In this case one obtains ring isomorphisms $$ M_{n+1}(R) \cong \mathrm{End}_R(R^{n+1}) \cong \mathrm{End}_R(R^n) \cong M_n(R). $$ The ring theorists provide us with examples of such rings. In fact, for any positive integers $n < m$, Leavitt gives a finitely generated ring $L_{n,m}$ for which there is a left $L_{n,m}$-module isomorphism $L_{n,m}^n \cong L_{n,m}^m$ and, consequently, a ring isomorphism $M_n(L_{n,m}) \cong M_m(L_{n,m})$.