Alternative proof of $I$ maximal implies $I$ prime The proof I know to show maximal implies prime for an ideal $I$ in a commutative ring $R$ goes as follows:
$I$ maximal $\iff$ $R/I$ is a field $\implies$ $R/I$ is an integral domain $\iff$ $I$ is prime.
Surely, it has to be possible to show the same as follows:
Let $I$ be maximal and $ab \in I$. Assume $a \notin I$. Then $I$ is properly contained in $I + (a)$ and hence $1 \in I + (a)$ that is, there exists $r \in R$ and $i \in I$ such that $i + ra = 1$. Similarly, one can find $i^\prime$ and $r^\prime$ such that $i^\prime + r^\prime b = 1$. So we get that $ra - r^\prime b \in I$. 
Now I'm stuck. How can I deduce that either $a$ or $b$ have to be in $I$?

Your goal is to prove that, when $I$ is maximal, $ab\in I$ implies that either $a\in I$ or $b\in I$. Note that this is equivalent to proving that $ab\in I$ and $a\notin I$ implies $b\in I$. 
If $ab\in I$ and $a\notin I$, then $I+(a)=R$ and hence there is some $r\in R$ and $i\in I$ such that $i+ra=1$ (you've already noticed this). Multiplying by $b$,
$$ib+rab=b.$$
$I$ is an ideal and $i\in I$, so that $ib\in I$, and we also have $ab\in I$ by assumption. Therefore, we can conclude that  $ib+rab=b\in I$.
