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?

You won't necessarily get a field in the quotient without commutativity, but you have a decent notion, nonetheless.
The rightmost equivalence is just the definition of simple ring.
If $I$ isn't maximal, then there is a proper ideal $J$ of $R$ with $I\subsetneq J$. Show that $J/I$ is a non-trivial ideal of $R/I$. Thus, simpleness of $R/I$ implies maximality of $I$.
On the other hand, suppose that $R/I$ isn't simple, so that there is a non-trivial ideal $\overline J$ of $R/I$. Let $J$ be the preimage of $\overline J$ under the quotient map $R\to R/I$, and show that $J$ is a proper ideal of $R$ containing $I$, so that $I$ is not maximal.
