Assume that $\mathcal{O} = \mathbb{Z}[x]/(q(x))$ is the ring of integers of a number field $K$. One can search for a prime $p$ which is not inert in $\mathcal{O}$ and such that there is a principal prime ideal above $p$. I think such primes exist because the density of principal prime ideals is $1/h$, where $h$ is the class number of $\mathcal{O}$ (see [this question][1]), and because the prime ideals of $\mathcal{O}$ lying above an inert prime have density 0 (because their norm is $p^{[K:\mathbb{Q}]}$).

Write $p \mathcal{O} = \mathcal{P} \cdot I$ where $\mathcal{P} = (a)$ is a prime ideal, $I = (b) \neq \mathcal{O}$ is coprime to $\mathcal{P}$ and $p=ab$ in $\mathcal{O}$. Then you will have $p=a(x)b(x)+q(x)r(x)$ for some $a(x), b(x), r(x) \in \mathbb{Z}[x]$, and Hensel's lemma will work because $\langle a(x), b(x), q(x) \rangle = \mathbb{Z}[x]$. It will give a factorisation of $p$ in $\mathbb{Z}[x]/(q(x))^n$ for every $n \geq 2$.


  [1]: https://mathoverflow.net/questions/375142/how-many-non-principal-prime-ideals-does-a-number-field-contain