Does every noninvertible element of a commutative ring lie in a proper maximal ideal?

More formally stated:

Prove that if $R$ is a commutative ring with $1$, then every element of $R$ that is not invertible is contained in a proper maximal ideal.

I know I have to assume Zorn's Lemma, but I don't see how non-invertible elements must lie in a proper maximal Ideal. Any hints?

This follows from showing that noninvertible elements lie in some proper ideal and that all proper ideals are contained in maximal ideals. In this vein, what is an ideal containing noninvertible $x$? It is proper? Now, consider the set of all ideals containing this ideal. How might you express the supremum of a chain $I_0\subset I_1\subset\dots I_n\subset\dots$? – peoplepower Jan 23 '13 at 8:50
It is well known that every proper ideal is contained in a maximal ideal. If $a$ is a noninvertible element, then the generated ideal $(a)$ is not the whole ring. If it were, then $1\in (a)$, implying $ab=1$ for some $b$, a contradiction. As a proper ideal, $(a)$, and hence $a$, must then be contained in a maximal ideal.