In light of Laurent Moret-Bailly's comment under Martin's answer, let's add the simple argument that finitely generated rings have prime and even maximal ideals, without appealing to anything like a choice principle.
If $R$ is finitely generated (a quotient of some $\mathbb{Z}[x_1, \ldots, x_n]$), then it is countable. Enumerate its elements: $a_1, a_2, \ldots$. Form ideals $P_n$ by recursion: define $P_0 = \{0\}$, and given $P_{n-1}$, define $P_n = P_{n-1} + \langle a_n\rangle$ if this doesn't contain $1$, otherwise define $P_n = P_{n-1}$. The union of the $P_n$ is then a maximal ideal: it is proper because $1$ doesn't belong to any $P_n$. And if $a \notin P$ for some $a = a_n$, then $a_n \notin P_n$, meaning that $1 \in P_{n-1} + \langle a_n\rangle$ according to how $P_n$ was defined, and then $1 \in P + \langle a\rangle$. This proves maximality.