It seems intuitive that if an irreducible polynomial $q\in\mathbb Z[X]$ has no primes in it's image, i.e. $n\in\mathbb Z\implies q(n)\notin \mathbb P$, then there is a non unit integer $m\in\mathbb Z$ such that $m|q(n)$ for all $n\in \mathbb Z$. But how to prove it?
Example: $q(x)=4x^5-7x^4-4x^3-4x^2+7x+8$ is irreducible and $2|q(n)$ all $n$.
I didn't have a single idea how to manage this, but I had a feeling that a theorem or conjecture was appropriate, but I had no key word, and thats the reason why I tagged reference-request
(I'll assume by no primes you mean in absolute value (allowing negative primes), otherwise $-x^2-1$ would represent no prime and yet the statement does not hold.)
This statement is equivalent to Bunyakovsky conjecture, an open problem. It states that an irreducible polynomial $f\in \mathbb{Z}[x]$ with positive leading coefficient and with $\gcd(f(1),f(2),\dots)=1$, represents infinitely many primes.
The positive leading coefficient is only technicality, if we allow negative primes $-p$, we can drop that condition. More importantly, you can replace the "infinitely many primes" with "at least one prime", and it will still be as strong as the original conjecture! (This post explains the equivalency On a (possible?) equivalence of Bunyakovsky conjecture, the trick is to repeatedly use the statement to construct additional primes).
So taking the contrapositive of the Bunyakovsky conjecture in this form, we get: if $f\in \mathbb{Z}[x]$ represents no primes (whether negative or positive), then it is either reducible or $\gcd(f(1),f(2),\dots)>1$. Since you assume the polynomial is irreducible, the conclusion of the implication simplifies to give your statement.