The nil-radical is an intersection of all prime ideals proof

Every proof I found online made the same implications.
Take one for example:
https://artofproblemsolving.com/wiki/index.php?title=Nilradical

I'm quoting the relevant part, which confuses me:

"To show the converse, it suffices to show that for any non-nilpotent element $a$, there is some prime ideal that does not contain $a$.

So suppose that $a$ is an element of $A$ that is not nilpotent. Let $S$ be the set of ideals of $A$ that do not contain any element of the form $a^n$. Since $(0) \in S$, $S$ is not empty; then by Zorn's Lemma, $S$ has a maximal element $\mathfrak{m}$.

It suffices to show that $\mathfrak{m}$ is a prime ideal. Indeed, suppose otherwise; then there exist elements $x,y \notin \mathfrak{m}$ for which $xy \in \mathfrak{m}$. Then the set of elements $z$ for which $xz \in \mathfrak{m}$ is evidently an ideal of $A$ that properly contains $\mathfrak{m}$; it therefore contains $a^n$, for some integer $n$. By similar reasoning, the set of elements $z$ for which $a^n$ $z \in \mathfrak{m}$ is an ideal that properly contains $\mathfrak{m}$, so this set contains $a^m$, for some integer $m$. Then $a^{n+m} \in \mathfrak{m}$, a contradiction.

Therefore $\mathfrak{m}$ is a prime ideal that does not contain $a$."

And then they conclude that if $a$ is not in the maximal element in $A$, and it is indeed prime, then $a$ is not in the intersection of all prime ideals, but who said there isn't some other prime ideal $P$ that is not in $A$?

Instead of commenting on this proof, which is too complicated in my opinion, here is a fast proof.

Assume that $a$ is not nilpotent. Then the localization $A_a$ is non-zero (otherwise $1/1=0$ in $A_a$, which would mean $a^n \cdot 1 =0$ for some $n$). Hence, it has a prime ideal $\mathfrak{p}$. The preimage in $A$ is a prime which doesn't contain $a$ (otherwise $\mathfrak{p}$ would contain a unit). Hence, $a$ is not contained in the intersection of all prime ideals of $A$.