There does not exist a onto ring homomorphism from $M_{n+1 \times n+1}(\mathbb F) \to M_{n \times n}(\mathbb F) $ for any field $\mathbb F.$

Let $A\in M_{n+1\times n+1}(\mathbb{F})$ be a nilpotent matrix such that $A^n\neq 0$, but $A^{n+1}=0$.

Since $A^{n+1}=0$ then $0=f(A^{n+1})=f(A)^{n+1}$.

Thus, $f(A)\in M_{n\times n}(\mathbb{F}) $ is also nilpotent. Therefore, $f(A)^{n}=0$. Hence $f(A^n)=0$.

Note that $A^n\neq 0$, so $\ker(f)$ is non trivial.

Since $\ker(f)$ is an ideal of $M_{n+1\times n+1}(\mathbb{F})$ and the only non null ideal of $M_{n+1\times n+1}(\mathbb{F})$ is itself then $f\equiv 0$.