Is it possible to classify all cyclic $\mathbb{Z}/4$-algebras, i.e. the regular quotients of $\mathbb{Z}/4 [X]$? A typical example is $\mathbb{Z}/4 [X] / \langle X^n , 2 X^k \rangle$. For my purposes it is not necessarily important to decide which quotients are isomorphic (this is already unclear to me for $\mathbb{Z}/2[X]$). So an almost identical question is the following:

Is it possible to classify all ideals of $\mathbb{Z}/4 [X]$? Can we bound the number of generators? Or is this classification not feasible?

It is known that every $\mathbb{Z}/4$-module (also without any finiteness assumptions) is a direct sum of copies of $\mathbb{Z}/2$ and $\mathbb{Z}/4$. This should be useful.

Partial results are also helpful for me. Instead of $\mathbb{Z}/4$ it should be possible to take $R/p^2$ for any principal ideal domain $R$ and a prime element $p \in R$.

*Edit.* Here is how I understand YCor's proof that every ideal of $R/p^2 [X]$ can be generated by two elements; the proof works for every commutative ring $A$ with an element $p \in A$ such that $p^2=0$ and $A/pA$ is a principal ideal ring. Let $I \subseteq A$ be an ideal. Consider the $A$-module $I \cap p A$. Since it is killed by $p$, we may view it as an $A/pA$-module. But $A/pA$ is a principal ideal ring, and $pA$ is a cyclic $A/pA$-module. Hence, $I \cap pA$ is a cyclic $A/pA$-module, i.e. a cyclic $A$-module. On the other hand, $I/(I \cap pA) \cong (I + pA)/pA$ is an $A$-submodule of $A/pA$, and hence cyclic too. Hence, $I$ is generated by a generator of $I \cap pA$ together with any preimage of a generator of $(I+pA)/pA$.