Proving a ring is Noetherian when all maximal ideals are principal generated by idempotents

Let $\mathfrak m = (r)$ be maximal. Note $r^2 = r$ can be written as $r(1 - r) = 0$ so every prime $\mathfrak p$ either contains $r$ or $1 - r$. As $\mathfrak m$ does not contain $1 - r$ this means every prime $\mathfrak p$ either equals $\mathfrak m$ or is not contained in $\mathfrak m$. In other words, all primes are maximal because they must equal the maximal ideal that contains them.

The maximal ideals are finitely generated by hypothesis.