Tajland, 2014

First, we note that the constant polynomials $P(x) \equiv 1$ and $P(x) \equiv -1$ satisfy the above divisibility condition. We show that these two polynomials are the only ones satisfying the divisibility condition.

Suppose that the polynomial $P(x)$ with the integral coefficients satisfying the above divisibility condition and $P(x) \neq 1$ and $P(x) \neq -1$. Note that $P(x) \neq 0$. If $P(\mathbb{Z}^+) \subseteq \{-1, 0, 1\}$, then there is a $j \in \{-1, 0, 1\}$ such that $P(x) - j$ has infinitely many zeros, a contradiction. So there is a positive integer $n_0$ such that $|P(n_0)| > 1$. Therefore, there exists a prime number $q$ such that $q \nmid P(n_0)$ and hence $q \nmid 2557^{n_0} + (213 \times 2014)$. Thus, $q$ is odd and $q \neq 2557$ ($2557$ is prime). Moreover, note that $P(n_0+q) \equiv P(n_0) \equiv 0 \pmod{q}$. Since $P(n_0+q) \equiv 2557^{n_0+q} + (213 \times 2014) \pmod{q}$, and $P(n_0) \equiv 2557^{n_0} + (213 \times 2014)$, $q \equiv 2557^{n_0} + (213 \times 2014) \pmod{q}$.
$$2557^{n_0+q} + (213 \times 2014) \equiv 0 \equiv 2557^{n_0} + (213 \times 2014) \pmod{q}.$$
Thus, $2557^{n_0+q} \equiv 2557^{n_0} \pmod{q}$. Since $(q, 2557) = 1$, $2557^q \equiv 1 \pmod{q}$. By Fermat's little theorem, $1 \equiv 2557^q \equiv 2557 \pmod{q}$. So $q \nmid 2556$. But $q$ is odd, $q \in \{3, 71\}$ which implies that $q \nmid (213 \times 2014)$. Since $q \nmid 2557^{n_0} + (213 \times 2014)$, $q \nmid 2557^{n_0}$ and hence $q = 2557$, a contradiction. So we are done. $\square$