**Proposition:** Let $(R, m)$ be a local ring. If $m$ is nilpotent, i.e. $m^n = 0$ for some $n$, then $m$ is the only prime ideal of $R$.
Proof: Let $p$ be any prime ideal of $R$, then $m^n = 0 \subseteq p \implies m \subseteq p$, so $m = p$ (since $m$ is maximal).
Now use the correspondence between primes sitting between $I$ and $p$ (i.e. $I \subseteq q \subseteq p$), and primes of $R_p/I_p$.