An ideal whose radical is maximal is primary 
I've got to prove that an ideal $Q$ whose radical is a maximal ideal is a primary ideal. That is, I want to prove that if $xy\in Q$, then $x\in Q$ or $y^n\in Q$ for some $n>0$. 

I've been trying for a while and I'm not sure where to begin. All I've got is that if $\text{Rad}(Q)$ is maximal and by definition it's the intersection of all prime ideals $P_i$ containing $Q$, then it must be equal to each of these $P_i$. So there is only 1 prime ideal containing $Q$, namely $\text{Rad}(Q)$.
Could anyone point me in the right direction? Thanks for any replies.

Hints: 
Reflect the problem in the quotient ring $R/Q$. Prove (if you don't already believe) that the definition of $Q$ being primary is equivalent to $R/Q$ satisfying "$ab=0$ implies $a=0$ or $b$ is nilpotent."
The condition that $\sqrt{Q}$ is maximal implies that $R/Q$ has exactly one maximal ideal, which is $\sqrt{Q}$. Thus it's a local ring.
Let $ab=0$ in $R/Q$ and suppose $b\notin \sqrt{Q}$. What can you say that is special about $b$ if it is outside the maximal ideal of a local ring? Conclude that $a=0$.
