Maximal \(\implies\) prime, but generally not the converse.
Solution
Assume rings are commutative with identity.
1. Every maximal ideal is prime.
Let \(\mathfrak m\) be maximal. Then \(R/\mathfrak m\) is a field, hence an integral domain. An ideal is prime exactly when its quotient ring is an integral domain. Therefore \(\mathfrak m\) is prime.
2. The converse fails in general.
In \(\ZZ\), the zero ideal \((0)\) is prime because \(\ZZ\) is an integral domain. But \((0)\) is not maximal, since \[ (0)\subsetneq (2)\subsetneq \ZZ. \] Thus a prime ideal need not be maximal.