I think that I have a proof, but it seems much simpler than all proofs that I can find on the internet. Hence I suppose that there must be a mistake in my proof.
The commutative ring $R$ is Noetherian if and only if each ideal $I<R$ is finitely generated as an $R$-module (this is easy to prove).
Suppose that $I<R$ a nonzero ideal for $R$ a commutative Artinian ring. Now there must exist a pair $(j_1,J_1)$ such that $J_1<I$ strictly and $J_1+j_1R=I$. Inductively we construct a strictly descending chain which must terminate as $R$ is Artinian, say $$I>J_1>J_2>...>J_n=0.$$ But then $I=(j_1,j_2,...,j_n)$, hence finitely generated. As $I$ was chosen arbitrarily all ideals are finitely generated and $R$ must be Noetherian.
Edit: I think the difficulty is in the existence of the pair $(j_1,J_1)$, is there any way to prove that such a pair does exist or that the idea is entirely wrong?