Let $R$ be a ring which is not a PID. The set $S$ of all ideals of $R$ which are not principal has a maximal element with respect to the inclusion (Zorn's Lemma), I already proved this! Now, I need to show that if $P$ is a maximal element of $S$, then $P$ is prime.
I tried by contradiction: suppose $a,b \not\in P$, but $ab \in P$. So $P+(a)$ and $P+(b)$ are not in $S$, and therefore they are principal ideals, say $P+(a) = (x)$ and $P+(b) = (y)$. Now, I just need to show that $P$ is principal. Any hint will be helpful.
Thus, I will conclude that for any ring, if all prime ideals are principal, then any ideal is principal.
See this link (includes proofs). Two steps:
The family of principal ideals of a ring is an Oka family.
If $\cal F$ is an Oka family of ideals, then any maximal element of the complement of $\cal F$ is prime.
In ring $R$, a set of ideals $\cal F$ is an Oka family if $R\in \cal F$ and whenever $I$ is an ideal such that $(I:a)\in \cal F$ and $(I,a)\in \cal F$ for some $a\in R$, then $I \in \cal F$.
Pedro Tamaroff-suggested other good link is here. Also the ones pointed out by Bill Dubuque in an MSE answer (for another standard Oka family, ideals that do not intersect a given multiplicative subset in the ring) here.