A regular element exists by prime avoidance

An element of $R$ is a zero divisor on $M$ exactly when it belongs to an associated prime: one direction follows from an associated element, and the other from A zero divisor is contained in an associated prime under the stated Choice hypothesis. Thus an $M$-regular element of $I$ exists exactly when $I$ is not contained in the union of the associated primes.

The associated-prime set is finite by Finite modules over Noetherian rings have finitely many associated primes under the stated Choice hypothesis. It is nonempty because $0$ is a zero divisor on $M\mathrm{\ne }0$ and step 1.1 places every zero divisor in an associated prime. Thus An ideal contained in a finite union of prime ideals lies in one of them applies and says that $I$ is contained in that union exactly when it is contained in one member. Combining this with step 1.1 produces a nonzerodivisor $x\in I$ exactly under the displayed condition; $M\mathrm{/}xM\mathrm{\ne }0$ because $xM=M$ would imply $IM=M$. Thus $x$ is regular in the adopted sense, proving both directions.