Let $f: R \to S$ be a surjective ring homomorphism.
Let $M \subset S$ be maximal, and let $f^{-1}(M) \subset I$ for some ideal $I \subset R$.
Then $M=f(f^{-1}(M)) \subset f(I)$ since $f$ is surjective.
Since $M$ is maximal in $S$, then either $f(I)=M$ or $f(I)=S$.
If $f(I)=M$, then $I \subset f^{-1}(f(I)) =f^{-1}(M)$, hence $I=f^{-1}(M)$.
Now, I'm having trouble showing that if $f(I)=S$, then $I=R$.