Let $R$ be a ring with $1$. a nonzero proper ideal $I$ of $R$ is a maximal ideal iff the $R/I$ is a simple ring. Let $R$ be a ring with $1$. Prove that a nonzero proper ideal $I$ of $R$ is a maximal ideal if and only if the quotient ring $R/I$ is a simple ring.

My attempt:-
$I$ is maximal $\iff$ $R/I$ is a field. $\iff$ $R/I$ has no non-trivial ideals $\iff$ $R/I$ is simple.
Is it correct?

Cameron's answer will give you all you need, but I'll sketch a proof of $I$ maximal $\implies R/I$ simple that doesn't involve contradiction, just in case you're interested in such things.
Suppose $I$ is maximal, let $\pi : R\twoheadrightarrow R/I$ be the natural projection, and let $J$ be a nonzero ideal of $R/I$. Then $\pi^{-1}(J)\subseteq R$ is an ideal of $R$ (strictly) containing $I$, so that $\pi^{-1}(J) = R$. However, this implies that $J = R/I$, so that $R/I$ is simple.
