I have this rough idea: if $\mathrm{rad}(q)$ is maximal, then $R/\mathrm{rad}(q)$ is a field. Consider $xy \in\mathrm{rad}(q)$, so $(x+\mathrm{rad}(q))(y+\mathrm{rad}(q))=\mathrm{rad}(q)$ is the additive identity. Thus either $x\in\mathrm{rad}(q)$ or $y\in\mathrm{rad}(q)$.
Would that be enough? If $R$ is commutative, would that be enough if I say that either of them must be in $\mathrm{rad}(q)$?
Consider $xy \in\mathrm{rad}(q)$ so $(x+\mathrm{rad}(q))(y+\mathrm{rad}(q))=\mathrm{rad}(q)$
Why only in $\mathrm{rad}(q)$? In your (title) question you are hypothesizing that $xy\in q$. No matter, we will see that it doesn't matter much shortly. (Incidentally, it would be better to write the right hand side as an actual coset: $(x+\mathrm{rad}(q))(y+\mathrm{rad}(q))=0+\mathrm{rad}(q)$.)
Thus either $x\in\mathrm{rad}(q)$ or $y\in\mathrm{rad}(q)$
This is true enough. Consequently, there is a positive power of $x$ in $q$ or a positive power of $y$ in $q$. Unfortunately there is a big problem: that isn't enough to conclude that $q$ is primary. So the approach does not get you to where you want to be.
You could go a little deeper and observe something about maximal ideals containing $q$. Or you could perform the most likely computation.
