If you do not allow the full rational field, you expect to lose the Euclidean property. That is, $\mathbb Z[x]$ is not a PID or a Euclidean domain.
A Euclidean domain $R$ is when we have some sort of degree function, Rotman calls it $\partial,$ with $\partial: R \setminus \{0\} \rightarrow \mathbb N,$ such that $\partial(f) \leq \partial(fg) $ for all nonzero $f,g \in R,$ and as soon as $f \neq 0$ we can always write $ g = q f + r,$ with either $r=0$ or $\partial(r) < \partial(f).$
Anyway, in $\mathbb Z[x]$ we take $\partial$ to be the degree of a polynomial. Suppose we then take $g=x$ and $f=2.$ We are asking about writing $ x = q \; 2 + r $ with either $r=0$ or $\partial(r) < \partial(f) = \partial(2) =0.$ The latter is impossible, so we need $r=0$ and $x = 2 q$ for some polynomial $q.$ This is fine in $\mathbb Q[x],$ but the evident solution $q = x/2$ is not a legal element of $\mathbb Z[x].$ We could be more formal about proving there is no $q,$ but in the end $\mathbb Z[x]$ is not a Euclidean domain.
Meanwhile, in case you wonder about some exotic degree map that makes $\mathbb Z[x]$ into a Euclidean domain, be aware that $\mathbb Z[x]$ is not a PID either, which rules out that strategy. Indeed, the ideal generated by $(2,x)$ is not principal.