Maximal ideal not containing the set of powers of an element is prime

In the midst of attempting to prove that for a commutative ring $A$ with identity, and an ideal $I$ of $A$, $I = \operatorname{rad}(I)$, where $\operatorname{rad}(I) = \{x: x^m \in I, m >0\}$, implies that $I$ is an intersection of prime ideals, I've run into a little bit of confusion.

An answer to another question on this site, this one to be exact, indicates a strategy that I am attempting to use. I know that if $x$ is an element of $A\setminus I$, then so too is $x^m$ for any positive integer $m$. I use this fact to consider a subset $X = \{x^m : m > 0\}$, which is closed under multiplication, and use Zorn's lemma to find an ideal of $A$ containing $I$ maximal with respect to the constraint that said ideal, $J$, is disjoint from X.

I want to prove $J$ is prime by contradiction. I suppose that $J$ were not prime; this also means that $J$ is not maximal in $A$. Then there would exist two elements, $f$ and $g$, of $A$ so that $fg$ is in $J$, but neither $f$ nor $g$ is in $J$. Following the prompting of the aforementioned answer, I know that $J + \langle a\rangle$ and $J + \langle b\rangle$ would be ideals strictly containing $J$, and since J is the maximal ideal disjoint from $X$, both $J + \langle a\rangle$ and $J + \langle b\rangle$ contain at least some elements of $X$ (I am tempted to say all of them, but I'm not entirely sure that's true, in case for example $\langle a\rangle$ started containing powers of $x$ at $x^2$, in which case it isn't necessarily true that $\langle a\rangle$ contains $x$ itself), and so both $\langle a\rangle$ and $\langle b\rangle$ must themselves contain elements of $X$.

And this is where I am stuck. I feel like there is something I should know about multiplicative subsets of rings that would help me arrive at the desired contradiction, but I can't remember what it might be.

This is a standard (and important) result, that also has a standard (and important) proof. The point is that for any $a, b \in A$, if $ab \in J$, then $(J + (a))(J + (b)) = J^2 + J(a) + J(b) + (ab) \subseteq J$. Since $J$ was maximal with respect to being disjoint from $X$, both $J + (a)$, $J + (b)$ contain elements of $X$, and $X$ is multiplicative, which gives a contradiction.

By the way, your observation that $J + (a)$ need not contain all of $X$ is correct, for precisely the reason you stated.