Suppose $M,N$ are two non-isomorphic simple $R$-modules. For $m,n\geq1$, is it true that $$ \text{Hom}_R(M^{\oplus m},N^{\oplus n})\cong\hat{0}\,? $$
I think it's true by Schur's Lemma.
However, I don't think $M^{\oplus m}$ will be simple in general, e.g. take $M:=R:=\mathbb{R}$ and $m:=2$, then I think $\mathbb{R}\oplus\hat{0}$ is a non-trivial left ideal of $M^{\oplus m}$, so it is a non-trivial submodule of $M^{\oplus m}$.
I think the reason is that $$ \text{Hom}_R(M^{\oplus m},N^{\oplus n})\cong[\text{Hom}_R(M,N)]^{\oplus mn}, $$ where Schur's Lemma applies to each term in the direct sum of the RHS.
The isomorphism of Hom group is already very good, but if you want to you can attack it this way:
If $f:M^m\to N^n$ was a nonzero homomorphism, then $f$ must be nonzero on one of the copies of $M$, and the restriction to this copy creates an injection of $M$ into $N^n$. I'll recycle the letter and call the image of this injection $M$ again.
Since $N^n$ is semisimple, $M\oplus K=N^n$ for some submodule $K$ of $N^n$. We can refine this decomposition by decomposing $K$ into simple submodules. By the Krull-Schmidt theorem, this refinement and $N^n$ are equivalent, so that they have the same number of terms and after some permutation, the components are isomorphic. But this would mean $M$ is isomorphic to one of the copies of $N$ in the decomposition $N^n$.
This proves the contrapositive of the desired statement.