It is a well-known fact that in ZF, the axiom of choice is equivalent to the statement that every commutative ring has a maximal ideal. On the other hand, for Noetherian rings, this is not necessary (either in the sense that it only requires dependent choice or that it requires no choice at all, depending on your definition of Noetherian). It can also be shown (in ZFC) that every Artin ring is Noetherian, and the proof of this fact in Atiyah-MacDonald essentially relies on the following two results:

In an Artin ring, every prime ideal is maximal (proved without choice), and

The nilradical of a ring is the intersection of all of the prime ideals in a ring (proved with choice).

Result 2. is equivalent to the statement that every Artin ring contains a prime ideal. To see this, note simply that given a non-nilpotent element $f$ in a ring $A$, there is a prime ideal of A not containing $f$ if and only if $A_f$ has a prime ideal. According to this answer, the fact that every commutative ring has a prime ideal is equivalent in ZF to the Boolean prime ideal theorem, so at least full choice is not necessary. Can this be weakened for Artin rings? Alternatively, is there another proof that Artin $\Rightarrow$ Noetherian that uses a weaker form of the axiom of choice (or no form of it)? Thank you.