The idea is quite straightforward but the details are a little bit cumbersome.I'll just show the sketch of my solution.If $Q(x)$ is not in the form of $cP(x)^2$,where $c\in\mathbb{N^*}$ and $P\in\mathbb{Z}[x]$,we could find a sufficiently large prime number $p$ satisfied that $p\nmid (a+1)^{Q(1)}-a^{Q(1)}$ and $p\mid Q_{1}(n_{p}+1)$,but $p\nmid Q'_{1}(n_{p}+1)$ for some $n_p\in\mathbb{N^*}$,where $Q_{1}(x)$ is the squarefree part of $Q(x)$ which has degree $d\geq 1$.Then there exist $\alpha\in\mathbb{N^*}$ and $p^{2\alpha}\Vert Q_{1}(n_{p}+1)$.By Hensel Lemma and CRT we could find $n\equiv n_p\mod p^{2\alpha}$ and $n\equiv 1\mod p-1$ satisfied that $p^{2\alpha+1}\Vert Q_{1}(n+1)$ and $p\nmid (a+1)^{Q(n)}-a^{Q(n)}$,which gives a contradition.If $Q(x)=cP(x)^2$,then $c((a+1)^{Q(n)}-a^{Q(n)})$ is also a perfect square.Choose prime $q$ sufficiently large so that $\gcd (c,(a+1)^q-a^q)=1$ and $q\mid Q(m)$ for some $m\in\mathbb{N^*}$.Since $(a+1)^q-a^q$ isn't a perfect square,we could choose a prime factor $p$ satisfied that $p^{2\alpha-1}\Vert (a+1)^q-a^q$ for some $\alpha\in\mathbb{N^*}$.It's well-known that $q\mid p-1$ so $p$ is sufficiently large,hence there exist $s\in\mathbb{N^*}$ and $p\nmid Q(s)$.By CRT we could choose $n\equiv m\mod p-1$ and $n\equiv s\mod p$,so that $q\mid Q(n)$ and by LTE Lemma,$p^{2\alpha-1}\Vert (a+1)^{Q(n)}-a^{Q(n)}$,a contradiction.And the problem is done.

Solved with erkosfobiladol. \[Q(n+1)((a+1)^{Q(n)}-a^{Q(n)})\in \mathbb{Z}^2\]Claim: There are finitely many primes $p$ such that there exists $v_p(Q(l))\equiv 1(mod \ 2)$. Proof: We will show that $v_p(Q(l))\equiv 1(mod \ 2)$ implies $p|(a+1)^{Q(1)}-a^{Q(1)}$. Suppose that $p\not | (a+1)^{Q(1)}-a^{Q(1)}$ and $p^{2m+1}||Q(n+1)$. We see that $v_p(Q(n+p^{2m+2}k+1))=2m+1$ hence $p|(a+1)^{Q(n+p^{2m+2}k+1)}-a^{Q(n+p^{2m+2}k+1}$ or $p|(a+1)^{Q(n+k+1)}-a^{Q(n+k+1)}$. Pick $k\equiv -n(mod \ p-1)$ in order to get a contradiction.$\square$ Claim: $Q(x)=cR(x)^2$ for some $R\in \mathbb{Z}[X]$ and a squarefree constant $c$. Proof: Let $Q(x)=cA_1(x)^{t_1}\dots A_n(x)^{t_n}$ where $A_i'$s are irreducible over integers. By Bezout, let $U_{i,j}A_i+V_{i,j}A_j=C_{i,j}$ and $C=lcm(C_{i,j})$. Pick a prime $p|A_i(x)$ and $p\not | C$ which exists by Schur. Since $(A_i(x),A_i'(x))=1$, by "extended" Hensel (we utilise $A_i$ is irreducible), we can choose $v_p(A_i(x))\equiv 1(mod \ 2)$ and this yields $t_i$ is even. Applying this process for each $i$ gives the result.$\square$ \[c((a+1)^{cR(n)^2}-a^{cR(n)^2})\in \mathbb{Z}^2\]Suppose that $deg \ R\geq 1$. Notice that $(a+1)^{R(n)^2}-a^{R(n)^2}\equiv-1(mod \ 4)$ thus, it cannot be a perfect square. Let $v_p((a+1)^{R(n)^2}-a^{R(n)^2})$ be odd. By LTE, we observe the following equations, \[v_p(c((a+1)^{cR(n)^2}-a^{cR(n)^2}))=v_p(c)+v_p((a+1)^{cR(n)^2}-a^{cR(n)^2})=2v_p(c)+v_p((a+1)^{R(n)^2}-a^{R(n)^2})\]However, $2v_p(c)+v_p((a+1)^{R(n)^2}-a^{R(n)^2})$ is odd which gives a contradiction as desired.$\blacksquare$