Answer: $P(x) = a x + b$, where $a \neq 0$, $b$ are rational numbers.

Obviously, $P(x)$ is not constant, so $\deg P > 0$.

First of all we prove that the coefficients of $P(x)$ are rational numbers. Let $\deg P = n$, consider $n+1$ distinct rational numbers $r_1, r_2, \dots, r_{n+1}$ and rationals $x_1, x_2, \dots, x_{n+1}$ such that $P(x_i) = r_i, \forall 1 \le i \le n+1$. Then by the Lagrange's interpolation theorem we have
$$P(x) = \sum_{i=1}^{n+1} r_i \frac{\prod_{j \neq i} (x - x_j)}{\prod_{j \neq i} (x_i - x_j)}$$
and it easily follows all coefficients of the polynomial are rational numbers.

Multiplying by appropriate integer we can assume that $P(x)$ has integer coefficients (and, of course, it satisfies the condition of the problem).

Let $a_n$ be the leading coefficient of $P(x)$, w.l.o.g. $a_n > 0$ and let $a_0$ be the last coefficient of $P(x)$.

Consider numbers $r$ of the kind $r = p_1 + a_0$, where $p_1$ is a prime number. There exist an integer $p_r$ and a positive integer $q_r$ with $(p_r, q_r) = 1$ such that $P(p_r/q_r) = r$. Then it is easy to see that $p_r \mid r - a_0 = p_1$ and $q_r \mid a_n$, hence, $p_r \in \{-1, 1, -p_1, p_1\}$.

There are infinitely many different $r$, so taking them into consideration we obtain infinitely many different corresponding pairs $(p_r, q_r)$.

We state that among these pairs there are infinitely many $(p_r, q_r)$, for which $p_r \in \{-p_1, p_1\}$, since otherwise $|p_r/q_r| \le 1$, but $r$ can take very large values; then at some moment the equation $P(p_r/q_r) = r$ is impossible.

W.l.o.g. assume that there are infinitely many pairs $(p_r, q_r)$ with $p_r = p_1$ (if sign is negative solution is similar). Since the number of divisors of $a_n$ is finite, there are infinitely many prime numbers $p_1$ and some constant $d$ (the divisor of $a_n$) with $P(p_1/d) = p_1 + a_0$.

Then consider another polynomial $f(x) = P(x/d)$. So, there are infinitely many numbers $a \in \mathbb{Z}$ for which $f(a) = a + a_0$ and, therefore, $f(x) \equiv x + c$. Finally, we have that $P(x) = d x + c$ is a linear polynomial.