An easy example of a non-surjective ring morphism that sends prime ideals to prime ideals is the embedding of a field into any extensions field: the only prime ideal of the field is $(0)$, which of course is mapped to the only prime ideal of the larger field. For example, $\mathbb{Q}\hookrightarrow \mathbb{R}$.
Could a surjective ring homomorphism send a maximal ideal to a prime ideal that is not maximal?
No. If $f\colon R\to S$ is a surjective ring homomorphism, and $M$ is a maximal ideal of $R$, then either $f(M)$ is maximal in $S$, or $f(M)=S$. To see this, note that the Lattice Isomorphism Theorem establishes an inclusion-preserving correspondence between ideals of $S$ and ideals of $R$ that contain $\ker(f)$. If $M$ contains $\ker(f)$, then because it is maximal in $R$ it follows by this correspondence that its image is maximal in $S$. If $M$ does not contain $\ker(f)$, then since $M$ is maximal and $M\lt M+\ker(f)\leq R$ with $M+\ker(f)$ an ideal, it follows that $M+\ker(f)=R$. Thus, $f(M) = f(M+\ker(f)) = f(R) = S$. So the image of $M$ is either maximal or the whole ring; it cannot be a non-maximal prime, if $f$ is surjective.
Note that the assertion that a morphism that sends ideals to ideals must be surjective requires rings to be unital and maps to be unital (rings to have a multiplicative identity, and morphisms to send the multiplicative identity of the domain to the multiplicative identity of the image). A simple counterexample without this assumption is to take the zero map into a nontrivial ring.
