Denote $P(x)$ as $a_kx^k+a_{k-1}x^{k-1}+\dots+a_1x+a_0$ Claim : $a_0=0$ Proof. For each prime $p$ we choose prime $p_1 \not\equiv 1 \pmod{p}$ such prime exists by Dirichlet’s Consider $n=p_1^{p-2} p$ Then, $\sigma(n)=(1+p_1+\dots+p_1^{p-2})(1+p) = \left(\dfrac{p_1^{p-1}-1}{p_1-1}\right)(1+p)$ By Fermat’s theorem $\implies$ $p \mid p_1^{p-1}-1$ But since $p \nmid p_1-1$ Thus, $p \mid \sigma(n) \mid P(n)$ From $p \mid n \mid P(n)-a_0$ $\implies$ $p \mid a_0$ for each prime $p$ Hence, $a_0=0$ as desired $\square$ _____________________________________________________________________________________________________________________________ Then, $P(x)$ became $a_kx^k+a_{k-1}x^{k-1}+\dots+a_1x$ Now for each prime $p$ choose $n=p_1^{\phi(p^2)-1}p$ which $p_1 \not\equiv 1 \pmod{p}$ Then, $\sigma(n)=(1+p_1+\dots+p_1^{\phi(p^2)-1})(1+p)=\left( \dfrac{p_1^{\phi(p^2)}-1}{p_1-1} \right) (1+p)$ Analogously, $p^2 \mid \sigma(n) \mid P(n)$ Since $p^2 \mid n^2 \mid P(n)-a_1n \implies p^2 \mid a_1n \implies p \mid a_1$ for each prime $p$ Thus, $a_1=0$ Do this repeatly will concluded that $P \equiv 0$ is the only answer $\blacksquare$

Lovely problem, here is a sketch of my solution. We use the well known fact that for polynomials $A$ and $B$ if $A(x)$ divides $B(x)$ for infinitely many $x$, then it does for all $x$. (Proof: consider quotient and remainder.) Hence from $p$ and $p^2$ we get $x+1 | P(x)$ and $x^2+x+1 | P(x^2)$, and since $x^2+1$ and $x^2+x+1$ are coprime, we obtain that $(x^2+1)(x^2+x+1) | P(x^2)$ and hence $\deg P \geq 2$. Now, to generalize this, we work with $p^{2^k}$ where $p$ is prime. If one shows that $\sum_{i=0}^{2^k} x^i$ is coprime to each of $\sum_{i=0}^{2^{k-s}} x^{2^si}$ for $s=1,2,\ldots,k$ then it follows that the product of all these $k+1$ polynomials (of degree $k\cdot 2^k$) divides $P(x^{2^k})$ and thus $\deg P \geq k$ which is impossible for all $k$. The coprimality follows easily by showing that the two polynomials have no common complex root (of unity) - it could be the case that one can also give an inductive proof uaing Euclid's algorithm, but I have not bothered to check this.

Wait, doesn't just this work (seems simpler than the above ideas)? Pick a odd prime $p$. Take any prime $q>p$. We have that $q \mid p^{q-1}-1$ by FLT, so $q \mid \frac{p^{q-1}-1} {p-1}=\sigma(p^{q-2})$ as $\gcd(q, p-1)=1$. Thus, $q \mid P(p^{q-2})$ for any such $q$. Since $p^{q-2}\equiv \frac{1}{p} \pmod q$, let's define $Q(x)=x^{\deg(P)}P(\frac{1}{x})$. Observe that $Q(p)\equiv p^{\deg(P)}P(p^{q-2}) \equiv 0 \pmod q$, so $q \mid Q(p)$ for infinitely many $q$. Thus, $Q(p)=0$ for all odd primes $p$, i.e. $Q \equiv 0$. Thus, $P \equiv 0$, done.

Let $p$ be a prime and $n$ a positive integer. Then $(p^{n+1}-1) \mid (p-1)P(p^n)$. Using the same well known fact as in Post #3, we must have $$x^{n+1}-1 \mid (x-1)P(x^n)$$as polynomials. Now let $\zeta$ be any primitive $(n+1)$-th root of unity. We then have $(\zeta-1)P(\zeta^n)=0$ so $\zeta^n$ is a zero of $P$. Thus, $P$ has infinitely many zeroes so it must be a zero polynomial.

Solved with blueberryfaygo_55. Fix $p$ to be any prime number. Setting $n = p^k$ gives that $\frac{p^{k + 1} - 1}{p - 1}$ divides $P(p^k)$ for any positive integer $k$. Now let $q > p $ be any prime. Fix $k = q - 2$. We have that $q$ divides $p^{q - 1} - 1$ by FLT but not $p - 1$, so $q$ divides $P(p^{q - 2})$. However, $p^{q - 2} \equiv \frac 1p \pmod q$, so $q$ divides $P \left( \frac 1p \right)$, or in other words, $\frac{P \left( \frac 1p \right) }{q}$ is an integer . Since infinitely many such $q$ exist, we must have that $P \left( \frac 1p \right) = 0$. But this is true for all primes $p$, implying that $P$ has infinitely many roots, so $P$ is the zero polynomial, which obviously works.

Here is another cheap way: For fixed $k \in \mathbb{N}$ consider $n=kp$ with $p$ large. Then $\sigma(n)=\sigma(k)(p+1)$ so that $p+1 \mid P(kp)$ and hence $p+1 \mid P(-k)$. But since $k$ is fixed and $p$ is arbitrarily large, this implies that $P(-k)=0$. But as $k$ was arbitrary, this shows that $P$ has infinitely many roots, hence is the zero polynomial.

Similarly to Tintarn, working only with products of two distinct primes suffices. Indeed, $q+1 | (p+1)(q+1) = \sigma(pq) | P(pq)$ so for any fixed $p$ we have $x+1 | P(px)$ as polynomials; in particular $P(-p)=0$ for any prime $p$, therefore $P\equiv 0$.

Lemma: Consider two polynomials $A,B \in \mathbb Z[x]$ then if $A(n) \mid B(n)$ for infinitely many positive integers $n$, then $A(x) \mid B(x)$ as well. Proof: WLOG $A,B$ have a positive leading coefficient otherwise we consider their negatives. Then clearly $\text{deg} A \le \text{deg} B$ must be true otherwise, since we know that there exista a value $C$ such that for all $x>C$, both $A,B$ are strictly increasing we will notice that the growth of $A$ would be larger than $B$'s growth which at some point would mean $A(x)>B(x)$ for all $x>c_0$ and this is a contradiction to the assumption. Now consider the division algorithm $B(x)=Q(x) \cdot A(x)+R(x)$ for $Q,R \in \mathbb Z[x]$, with $\text{deg} R< \text{deg} A$ and $\text{deg} Q \le \text{deg} B$, so this will become $\frac{B(x)}{A(x)}=Q(x)+\frac{R(x)}{A(x)}$, this the implies that for such infinite positive integers $n$ we also have $\frac{R(n)}{A(n)} \in \mathbb Z$ but notice that at some point due to $A$'s growth we will have for some super large such $n$ that $-1<\frac{R(n)}{A(n)}<1$ since each time it'll get closer to zero, this is of course a contradiction to the fact that this quanty is an integer unless it's zero, which would mean that $Q$ has infinitely many large roots so it is the zero polynomial, therefore $A(x) \mid B(x)$ as desired. Back to the problem notice we have $p^n-1 \mid (p-1)P(p^{n-1})$ for all primes $p$ and a positive integer $n$ which now by the lemma gives us that $x^n-1 \mid (x-1)P(x^{n-1})$ therefore if we consider $\zeta$ as a n-th root of unity for $n \ge 1434$ and using the fact that $\zeta-1 \ne 0$ we get that $P(\zeta^{n-1})=0$ but this is also yet another n-th root of unity so all n-th roots of unity (other than $1$) for $n \ge 1434$ divide $P$ which means that $P$ has infinitely many roots. Therefore $P(x)=0$ must be true, and clearly satisfies the condition, thus we are done .