Let $R$ be a PID. Every proper ideal is a finite product of maximal ideals, which are uniquely determined up to order.
Here is a partial solution: Since $R$ is a PID, it must also be a UFD. If $I \subseteq R$ is a nontrivial proper ideal, then there exists $x \in R \setminus \{0\}$ that is not a unit, otherwise $I=R$. Since $x$ is a nonzero, nonunit, there exists irreducible elements $c_1,..,c_n$ such that $x=c_1...c_n$ and therefore $(x) = (c_1...c_n) = (c_1)...(c_n)$. Recall that $c_i$ is irreducible iff $(c_i)$ is maximal with respect to all principal ideals; but as they are the only ideals, $(c_i)$ is maximal in $R$.
As one may notice, I haven't dealt with uniqueness yet. I found this and this, but unfortunately uniqueness isn't discussed in either of these MSE posts. I have tried quite few different things (e.g., showing this element divides that element, and that element divides this element, etc.) but I haven't had much luck. I could use a hint.
EDIT:
Here is the best I can do: Suppose that $(x) = (p_1)...(p_m) = (p_1...p_m)$ Then $x \mid (p_1 ... p_m)$ and $(p_1 ... p_m) \mid x$ and therefore $p_i \mid x$ for every $i$. But I don't know how to conclude from this $x = p_1...p_m$.