$x$ not nilpotent implies that there is a prime ideal not containing $x$.

HINT $\ $ If $x$ isn't nilpotent then localize at the monoid generated by $x$ to deduce that that some prime ideal doesn't contain $x$.

Let $f$ not be nilpotent. Consider the set $\Sigma$ of ideals $I$ such that if $n > 0$ then $f^n \notin I$. It is non-empty because $(0) \in \Sigma$. Order $\Sigma$ by inclusion and by Zorn's lemma choose a maximal element $P$. Show that $P$ is prime: let $x,y \notin P$. Then the ideals $P+(x)$ and $P+(y)$ strictly contain $P$, and so er not in $\Sigma$. Thus $f^m \in P+(x)$ and $f^n \in P+(y)$ for some $m,n$ (by def of $\Sigma$). It follows that $f^{m+n} \in P+(xy)$, hence $P+(xy) \notin \Sigma$, that is $xy \notin P$, so $P$ is prime. Hence we have a prime ideal $P$ not containing $f$.

(the proof is really just copy from the proof in Atiyah-MacDonalds' book)