Yes, an integer polynomial has no integer cycle of length more than $2$.
Let's give a quick proof of this result. Let $P \in \mathbb{Z}[X]$, and $a_0$, ..., $a_{k-1} \in \mathbb{Z}$ such that
$$P(a_0)=a_1, \quad P(a_1)=a_2, \quad ... \quad P(a_{k-2})=a_{k-1}, \quad P(a_{k-1})=a_k$$
We define, for every $n \in \mathbb{N}$, $a_n = a_{(n \ \mathrm{mod}\ k)}$. We shall prove that either the sequence $(a_n)_{n \in \mathbb{N}}$ is constant, either it is $2-$periodic.
For each $i \in \mathbb{N}$, one has $a_{i+2}-a_{i+1} = P(a_{i+1})-P(a_{i}) = (a_{i+1}-a_i) R_i$ for a certain integer $R_i$, with $R_i=R_{(i \ \mathrm{mod}\ k)}$ since the sequence $(a_i)_{i \in \mathbb{N}}$ is $k-$periodic.
Thus, il follows by induction that for every $n \in \mathbb{N}$, one has
$$a_{n+k+1}-a_{n+k} = (a_{n+1}-a_n) \prod_{i=0}^{k-1} R_i$$
But one has $a_{n+k+1}-a_{n+k} = a_{n+1}-a_{n}$ so
$$(a_{n+1}-a_n) \prod_{i=0}^{k-1} R_i = a_{n+1}-a_n$$
so either $a_{n+1}=a_n$, either $\displaystyle \prod_{i=0}^{k-1} R_i = 1$.
The first case leads to the fact that $(a_n)_{n \in \mathbb{N}}$ is constant, so we can assume that $\displaystyle \prod_{i=0}^{k-1} R_i = 1$ : we deduce that for every $i \in \mathbb{N}$, $R_i = \pm 1$.
If one has $R_i=1$ for every $i \in \mathbb{N}$, then it means that $a_{i+2}-a_{i+1}=a_{i+1}-a_i$ : since the sequence $(a_n)_{n \in \mathbb{N}}$ is $k-periodic, then it implies that it is constant.
If one has $R_i=-1$ for an integer $i \in \mathbb{N}$, then for this $i$, one has $a_{i+2}-a_{i+1} = -(a_{i+1}-a_i)= a_i-a_{i+1}$ and hence, $a_{i+2}=a_i$ and the sequence is $2-$periodic.
