Subring of $\text{Mat}_n(Z_m)$ is commutative if $x^2=0 \implies x=0$.

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Let $A$ be subring of $\text{Mat}_n(Z_m)$. Suppose, for $x\in A$, $x^2=0$ implies $x=0$.
Claim A is commutative.

Attempt $A$ is finite ring, hence Artinian.
If it is possible to claim that Jacobson Radical is $0$, then by Weddenburg-Artin, result might follow.
(I did not finish the argument yet, I am not sure if this path is okay.)

I am not sure how to use the information $x^2=0 \implies x=0$.
Also, I am not sure where to use being subring of $\text{Mat}_n(Z_m)$ (except being finite).

A brief explanation might help a lot, thanks in advance.