this one is very easy: suppose deg(p(x))>=2. Suppose leading term is positive. Eventually the first term will 'overwhelm' the other terms, and similarly with the derivative. So we can find some integer q, s.t. p(x) is strictly increasing and p'(x) is also increasing for all x>q. But we have a contradiction in this because: draw lines corresponding to y=p(q)+1,y=p(q)+2,etc. The x-coordinates corresponding to these points are integers but they become closer and closer together, contradiction. a similar contradiction is reached if the leading term is negative. So it remains to consider when p(x) is linear or constant .

The solution above has main idea, but here is a better-written version. First, constant polynomials work, thus suppose $P(\cdot)$ is non-constant. Furthermore, notice we can take wlog the leading coefficient of $P$ to be positive. Now, as above, $P'$ is eventually positive, thus $P$ is eventually strictly increasing. Now, let $(N,\infty)$ be the interval on which $P$ is strictly increasing, and let $n\in(N,\infty)$ be an integer. Then, there is an $x_n$ such that $P(x_n)=n$ holds. The conditions of the problem imply $x_n\in\mathbb{Z}$. Now, there is another integer $x_{n+1}$, such that $P(x_{n+1})=n+1$. This implies, $x_{n+1}-x_n \mid P(x_{n+1})-P(x_n)=1$, and since we focus on the region on which $P$ is increasing, we get $x_{n+1}=x_n+1$. Similarly, get an $x_{n+2}$ such that $P(x_{n+2})=n+2$, and obtain that $x_{n+2}=x_n+2$. It therefore holds that, $P(x_n+\ell)=n+\ell = (n-x_n)+x_n+\ell$ for every non negative integer $\ell$, that is, $P(r)=r+C$ holds (for $C=n-x_n$ constant) infinitely often. However, this implies $P(r)-C-r$ has infinitely many roots, and thereby must itself be the zero polynomial. This yields, $P(x)=x+C$ where $C\in\mathbb{Z}$, in addition to constant solutions.