Let $I$ be such an ideal, and let $P$ be a minimal prime ideal over $I$. Then in $R/P$, every proper ideal is prime. Hence for any nonzero $a \in R/P$, $(a^2)$ is prime, so it contains $a$, so there is some $b$ such that $a^2b = a$; that is, $a(ab-1)=0$ and since $R/P$ is an integral domain, it follows that $ab=1$ and $a$ has an inverse. So that means that $R/P$ is a field; hence $P$ is maximal.
Now consider $R/I$. The nonzero proper ideals of $R/I$ correspond exactly to the proper ideals of $R$ strictly containing $I$, and they are all maximal. Use the result of your previous question to deduce that there are most two maximal ideals in $R/I$, and hence at most two proper ideals of $R$ strictly containing $I$.