Ugly! Love it! Let $m=2^k$. Suppose $\deg Q=d\ge2$. Then there are $C,c>0$ such that $Ck\ge\log_2(P(m))\ge f(P(m))=Q(f(m))=Q(k)\ge ck^d$ for all large enough $k$, so contradiction. Thus $\deg Q\le1$. Now let $P(x)=x^rR(x)$, where $R(0)\ne0$. Since $f$ is multiplicative, the equation becomes $f(R(m))=af(m)+b$. Suppose, $R$ is nonconstant. Take some large odd prime $q$, coprime with $R(0)$, such that $q\mid R(l)$ and $q\not\mid R'(l)$ for some $l\in\mathbb N$ (there is such, since $R$ is nonconstant). Take $s>2a+b$ and lift $l$ to $a+q^sn$, $n\in\mathbb N$, by Hensel. Obviously, $(a,q)=1$. (Next, I'll prove there are infinitely many semiprimes in $S=\{a+q^sn:n\in\mathbb N\}$ by elementary means. If Dirichlet theorem on primes in APs is allowed, skip the next part right to $p\in S$.) It's an elementary result that for every $h\ge2$ and every proper subgroup $H\subset G=(\mathbb Z/h)^\times$ there are infinitely many primes $p$ such that $[p]\in G\setminus H$. $G$ is cyclic generated by $g$ for $h=q^s$, so take $H=\{g^{2i}:i\in\mathbb Z\}$. Therefore, for every $a\in G$ there are (infinitely many) primes $p_1,p_2$ such that $[p_1p_2]=a$ in $G$. Take $m=p_1p_2$ and get $q^s\mid R(m)$, so $2a+b<s\le f(R(m))=af(m)+b=2a+b$, so contradiction. Therefore, $R$ is a constant and thus $P(x)=Rx^r$ and $Q(x)=rx+f(R)$.

Quick proof of mine: 1. $\deg Q \leq 1$ Proof. $n=p_1^{e_1}\cdots p_k^{e_k}\geq2^{f(n)}$ and $f(n)\leq \log_2{n}$. We put $m=2^k$ and $Q(k)=f(P(2^k))\leq\log_2Q(2^k)=O(k)$. 2. If $P$ is not constant, $P(0)=0$ Proof. Assume the contrary, and let $p_t$ be the $t$-th smallest prime. Let $|P(0)|<p_l$. We put $m=|P(0)|\cdot2\cdot3\cdot\cdots\cdot p_l\cdot2^k$. Then, regardless of $k$, the $p_i$-adic valuation of $P(m)$ is constant for $i=1,\ldots,i$. Thus for some constant $C$ (regarding $l$) $f(P(m))\leq C+\log_{p_{l+1}}P(m)$, bounding the left hand side, whilst in fact that $f(m)$ is linear so the right hand side is growing linearly. Thus for some large $l, k$ this yields a contradiction. If the right hand side is constant, clearly $P$ must be too. 3. $P$ is a monomial. Proof. Assume the contrary. Let $x^k\mid P(x)$ and $x^{k+1}\nmid P(x)$. Clearly $P(x)/x^k$ and $Q(x)-kx$ is a solution, and yet $P(x)/x^k$ is not constant and its constant term is nonzero, a contradiction to (2.). 4. Finding the solutions Now let $P(x)=ax^b$. Then $Q(x)=bx+f(a)$. Problem done. P.S. The only problem I managed to solve is this one, and just this made me into the top 16.

Acorn-SJ wrote: For each positive integer $n>1$, if $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$($p_i$ are pairwise different prime numbers and $\alpha_i$ are positive integers), define $f(n)$ as $\alpha_1+\alpha_2+\cdots+\alpha_k$. For $n=1$, let $f(1)=0$. Find all pairs of integer polynomials $P(x)$ and $Q(x)$ such that for any positive integer $m$, $f(P(m))=Q(f(m))$ holds. Let $P(x)=x^ah(x)$ with $(h(x),x)=1$ . Suposse that $h(x)$ is not constan then from Deleclert,Hensen lemma,Shur we can find a prime number $r$ such that $U_p(h(r))>Q(1)$ ($p$ is also prime) Then for $m=r$ in the equation we get a contradiction. So $P(x)=c*x^a$ and the equation became $af(m)+f(c)=Q(f(m))$ so we get that $Q(x)=ax+f(c)$