Let $A$ be a not necessarily commutative unital ring with a unique simple module (up to isomorphism). Let $\mathfrak m$ be the annihilator of this simple module, which is a two-sided ideal. We claim that $\mathfrak m$ is a maximal two-sided ideal. If $I$ is a maximal left ideal, then $A/I$ is a simple module and its annihilator is contained in $I$, since any annihilating element must kill $1+I$. If $J$ is a two-sided ideal contained in $I$, then $J$ must annihilate $A/I$, since if $x\in J, y\in A$, then $x(y+I)=xy+xI\subseteq I$, since $xy\in J\subseteq I$. Now, if $M$ is a maximal two-sided ideal (which exists by Zorn's Lemma), then there's a maximal left ideal $I$ containing $M$ (again by Zorn). Then, $R/I$ is simple and its annihilator is a two-sided ideal containing $M$ and thus equal to $M$, which also equals $\mathfrak m$ because there's a unique simple module. Hence, $\mathfrak m$ is the unique maximal two-sided ideal.
If $A$ is an Artinian ring, then $A/\mathfrak m$ is also an Artinian ring (since any infinite descending chain of left ideals in the quotient lifts to an infinite descending chain in $A$). Furthermore, $A/\mathfrak m$ is a simple ring since $\mathfrak m$ is a maximal two-sided ideal, so by Artin-Weddenburn, $A/\mathfrak m$ is isomorphic to a matrix algebra over a division ring. Is this true if we don't assume $A$ is Artinian?