If an integral domain $R$ doesn't satisfy the ascending chain condition for principal ideals, then factorization of an element of $R$ (into irreducible elements) might fail: if you choose the wrong elements, the factorization process might not terminate. However, can it be the case, that in some such domain, a factorization still exists for all elements (or at least for some element where the factorization can fail)? It could be that maybe if you choose the correct divisors then the factorization terminates.
The only example i know of a integral domain that doesn't satisfy ACCPI is $P/I$ where $P$ is the ring of polynomials of countable many variables $(x_k)$ over an integral domain $R$ and $I$ is the ideal generated by the relations $x_{k+1}^2 = x^k$. I thought of maybe trying to adjoin some new elements $y_k$ such that $x_k = y_k^{p(k)}$ where $p(k)$ is the kth prime starting at $3$ and hope those $y_k$'s turn out to be irreducible and that the result still doesn't satisfy ACCPI, as then an element $x_i$ would maybe have non-terminating factorization if you choose $x_k$'s to factorize it but would terminate if you factor by $y_i$. However this ring seems very complicated to work with and i don't have strong reasons to believe it will work.