$\rm 9=degP(Q(x))=degP(x)degQ(x)$.Also $\rm P(Q(k))=0,k=1,2,..,9$ so $\rm P(x)$ has 9 roots ,therefore $\rm degP(x)=9,degQ(x)=1$. We have that $\rm P(x)=m\prod_{k=1}^9 (x-Q(k)),m\in \mathbb{Z}$.Let $\rm Q(x)=ax+b,a,b\in \mathbb{Z}$. Then $\rm P(Q(x))=m\prod_{k=1}^9(Q(x)-Q(k))=ma^9\prod_{k=1}^9(x-k)$.So $\rm ma^9=1$. Case 1: $\rm m=1\Leftrightarrow a=1$ so we have the solution $\rm Q(x)=x+b,b\in \mathbb{Z},P(x)=\prod_{k=1}^9(x-Q(k))$. Case 2:$\rm m=-1\Leftrightarrow a=-1$ so we have the solution $\rm Q(x)=-x+b,b\in \mathbb{Z},P(x)=-\prod_{k=1}^9(x-Q(k))$.

Let $p,q$ and $a,b$ be the degrees and leading coeffitients of $P,Q,$ respectively. The given condition implies \[\begin{cases} pq=9\\ ab^{p}=1 \end{cases}\]so $a=b=\pm 1$ and $(p,q)=(1,9),(9,1),(3,3),$ so we consider 3 cases: $Case1.\;\;(p,q)=(1,9).$ Then $P(x)=ax+k$ for some $k\in \mathbb{Z}.$ So, solving for $Q,$ we obtain $$Q(x)=a(x-1)(x-2)\cdots (x-9)-ak$$(because $a^{2}=1$). $Case2.\;\;(p,q)=(9,1).$ Then $Q(x)=ax+m$ for some $m\in \mathbb{Z}.$ So we solve for $P,$ we obtain $$P(x)=P(Q(ax-am))=(ax-am-1)(ax-am-2)\cdots (ax-am-9)=a\prod_{j=1}^{9}(x-m-ja)$$So we got 4 families of solutions from these two cases. $Case3.\;\;(p,q)=(3,3).$ Let $r_{1},r_{2},r_{3}$ be the zeros of $P,$ then $Q(k)\in \{r_{1},r_{2},r_{3}\}\;\forall \;1\le k\le 9,$ so $r_{1},r_{2},r_{3}$ are pairwise distinct (otherwise at least 5 among the numbers $Q(1),Q(2),...,Q(9)$ are equal), moreover, each of the polynomials $Q-r_{1},Q-r_{2},Q-r_{3}$ will have 3 zeros from the set $\{1,2,3,...,9\}$. Let $x_{i},y_{i},z_{i}$ be the zeros of $Q-r_{i}$ for $i=1,2,3,$ then $(x_{1},y_{1},z_{1},x_{2},y_{2},z_{2},x_{3},y_{3},z_{3})$ is a permutation of $(1,2,3,4,5,6,7,8,9),$ and since the polynomials $Q-r_{1},Q-r_{2},Q-r_{3}$ agree on all coefficients except for the last one, we get from Vieta's relations: \[\begin{cases} x_{1}+y_{1}+z_{1}=x_{2}+y_{2}+z_{2}=x_{3}+y_{3}+z_{3}\\ x_{1}y_{1}+x_{1}z_{1}+y_{1}z_{1}=x_{2}y_{2}+x_{2}z_{2}+y_{2}z_{2}=x_{3}y_{3}+x_{3}z_{3}+y_{3}z_{3} \end{cases}\]hence, $$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=x_{2}^{2}+y_{2}^{2}+z_{2}^{2}=x_{3}^{2}+y_{3}^{2}+z_{3}^{2}=\frac{1^{2}+2^{2}+...+9^{2}}{3}=95$$say $z_{1}=9,$ then $x_{1}^{2}+y_{1}^{2}=14,$ which has no integer solutions! Therefore, there are only 4 solutions: \[\begin{cases} P(x)=x+k,\;\;Q(x)=(x-1)(x-2)\cdots (x-9)-k\\ P(x)=-x+k,\;\;Q(x)=-(x-1)(x-2)\cdots (x-9)+k\\ P(x)=(x-k-1)(x-k-2)\cdots (x-k-9),\;\;Q(x)=x+k\\ P(x)=-(x-k+1)(x-k+2)\cdots (x-k+9),\;\;Q(x)=-x+k \end{cases}\]where $k$ is an arbitrary integer.