Injective integer polynomial is injective modulo some prime

Consider $Q(x)=x(2x-1)(3x-1)$. This gives an injective map $\mathbb Z\to \mathbb Z$, because $n<m \implies Q(n)<Q(m)$. However, this $Q$ is not injective over $\mathbb Z/p\mathbb Z$ for any $p$ because $Q(x)=0$ has three solutions when $p\geq 5$ and two solutions when $p\in \{2,3\}$.

While the original question has been answered, there is a beautiful result of Fried in On a conjecture of Schur which is relevant to such questions. Suppose $Q$ is a polynomial in ${\Bbb Q}[x]$ such that for infinitely many primes the induced map from ${\Bbb Z}/p{\Bbb Z}$ to ${\Bbb Z}/p{\Bbb Z}$ is bijective. Then $Q$ must be of the form

(i) $Q(x) = ax^n + b$,

or

(ii) $Q(x) = T_n(x)$, where $T_n(x)$ denotes the $n$-th Chebyshev polynomial,

or

(iii) Compositions of functions of this type.

This established an old conjecture of Schur. See also the account Turnwald - On Schur's conjecture, which discusses the history of the problem, and gives a detailed proof, correcting inaccuracies in the earlier literature.