Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$?

According to the Amitsur-Levitzki theorem, $n \times n$ matrices over a commutative ring satisfy a polynomial identity of degree $2n$ and none of smaller degree. So there can be no injective ring homomorphism $M_{n+1}(R) \to M_n(R)$, which at least rules out the case when $R$ is a field.

Here is a slightly silly example of a non-trivial ring homomorphism $M_{n+1}(R)\rightarrow M_n(R)$ with $R$ noncommutative.

Let $R=\mathbb C \times M_2(\mathbb C)$. Then there is a ring homomorphism $M_2(R)\rightarrow M_1(R)$ sending the ideal $M_2(M_2(\mathbb C)) \subseteq M_2(R)$ to $0$ and $M_2(\mathbb C) \subseteq M_2(R)$ isomorphically to $0 \times M_2(\mathbb C)$.

There might be more interesting examples based on this idea.

If $R$ is a local ring, possibly non-commutative, then there is no non-trivial homomorphism. Let $B_k$ be the semigroup of $k\times k$-matrix units and $0$. It is well known every proper homomorphic image of $B_k$ collapses all elements. So if $M_{n+1}(R)\to M_n(R)$ is nontrivial it must not collapse $B_{n+1}$ (since $B_{n+1}$ spans $M_{n+1}(R)$). But then $B_{n+1}$ embeds in $End(R^n)$ as a semigroup with zero. So $End(R^n)$ contains $n+1$ orthogonal idempotents. But this implies $R^n$ is a direct sum of at least $n+1$ non-zero projective modules. But projective is free for local rings and local rings have invariant basis number. This is a contradiction.

Added. This argument works as long as R has the invariant basis number property for finitely generated free modules and finitely generated projective $R$-modules are free. In particular it applies to free algebras and firs (free ideal rings) if memory serves.