Maximal ideal contains a zero divisor 

Suppose $R$ is a commutative and unital ring. Let the ideal $I$ be maximal and $a,b$ be (nonzero) zero divisors in $R$.
Show that $ab = 0$ implies $a \in I$ or $b\in I$


We've only had a bit of exposure to ideals: we know that $I$ maximal $\to R/I$ field, a little about the Euclidean algorithm, and the definition of a PID.
I'm not sure how to approach this.  The problem seems simple and I'm probably just missing something.
Should I try assuming $a,b \notin I$ and try to derive a contradiction?

If $I$ is a maximal ideal then $I$ is a prime ideal. Indeed, let $ab \in I$ such that $a \notin I$. It suffices to show $b \in I$. Now, $I$ is a subring of $I+Ra$ and $I+Ra$ is an ideal, so $R=I+Ra$. Then there exists $m \in I$ and $r \in R$ satisfying $1=m+ra$. Thus, $b=bm+rab \in I+I \subset I$.
Now, the claim is that then $R$ is an integral domain if and only if $(0)$ is a prime ideal
Indeed, $(0)$ is a prime ideal if and only if $ab \in (0)$ implies  $a \in (0)$ or $b \in (0)$ if and only if $ab=0$ implies $a=0$ or $b=0$. 
Here, we have shown that if $I$ is maximal, then $I$ is a prime ideal, and further take $I=(0)=0.$ The result should now follow.
