The usual proof that irreducibles are prime in $\mathbb{Z}[x]$ uses Gauss's lemma to show that a polynomial with content 1 that is irreducible over $\mathbb{Z}[x]$ is also irreducible over $\mathbb{Q}[x]$. Then you can use that $\mathbb{Q}[x]$ is a PID to show that every ideal generated by an irreducible $p$ is maximal, from which it follows that $\mathbb{Z}[x]/\langle p \rangle$ is an ID and thus $p$ is prime in $\mathbb{Z}[x]$.
Is there a way to show that every irreducible is prime in $\mathbb{Z}[x]$ without "going" through $\mathbb{Q}$? You can show that $\mathbb{Z}[x]$ is an ID and has the ACCP directly (since it inherits these from $\mathbb{Z}$). But these are definitely not enough (indeed $\mathbb{Z}[\sqrt{-5}]$ also satisfies these properties but not all irreducibles are primes), so what makes $\mathbb{Z}[x]$ "special"?
Some added context: this comes from a homework question where I'm tasked with showing that $\mathbb{Z}[x]$ has the ACCP and that every irreducible is prime, in order to prove that it's an UFD. I know how to prove that irreducibles are prime by assuming that $\mathbb{Z}[x]$ is an UFD / by passing to $\mathbb{Q}[x]$ - I was wondering if one could avoid using unique factorisations in order to prove that irreducibles are prime.
Edit: The question this was flagged as being a duplicate of concerns itself with showing that factorial rings are AP-rings ("atoms are prime", so basically what I'm asking here). What I'm specifically asking is whether or not there is a way to prove that $\mathbb{Z}[x]$ is an AP-ring without involving $\mathbb{Q}[x]$, so unless somebody provides a proof of $\mathbb{Z}[x]$ being factorial that does not involve $\mathbb{Q}[x]$ / gives some heuristic reason as for why you're forced to pass to the field of fractions, I don't see how this question can possibly be a duplicate.
