Do we get this lucky property in Noetherian rings? One definition of a Noetherian ring is that every ideal is finitely generated. If $R$ is a Noetherian ring and $I$ and idempotent ideal, that is, $I^2=I$, would it be too far-fetched to guess that $I$ is generated by an idempotent element?

It is true, and a consequence of Nakayama's lemma: since $I$ is a finitely generated $R$-module such that $I\cdot I = I$, there is an $x \in I$ such that for all $y \in I$, $(1-x)y=0$, that is $y=xy$. Hence, $x$ is the required idempotent.