Yes. We take $P(x)=(x^3+3)(x^8-16)$. We now show that this polynomial indeed works. Step 1. Let $n=\prod_{1\le i\le N}p_i^{\alpha_i}$, where $p_1<\cdots<p_N$ are primes, and $\alpha_i$, $1\le i\le N$ are exponents. Note that by the Chinese remainder theorem, if$m_i$ are such that $p_i^{\alpha_i} \mid P(m_i)$, then taking $m\equiv m_i\pmod{p_i^{\alpha_i}}$, $1\le i\le N$, we deduce $n\mid P(n)$. Hence, it suffices to show this claim only for prime powers. Step 2. We now handle odd prime powers. For this, the gist is the polynomial $x^8-16=(x^2-2)(x^2+2)(x^4+4)=(x^2-2)(x^2+2)((x-1)^2+1)((x+1)^2+1)$. Note that at least one of $-2,2,-1$ are quadratic residues (modulo any odd prime $p$), thus for any odd $p$ prime, there is an $x$ such that $p\mid x^8-16$. Notice, furthermore, that $x\not\equiv 0\pmod{p}$ as $p>2$, hence by Hensel's lemma, we deduce for any prime $p>2$ and exponent $k$, there is an $x$ so that $p^k\mid x^8-16$. Thus the odd prime powers are handled. Step 3. We finally handle the powers of $2$. For this we use the guy $x^3+3$. We claim, for any $k$, there is an $x$ such that $2^k\mid x^3+3$. We show this as follows (I am sure this step can be compressed ---anyways---): consider $x^3+3$ modulo $2^k$, where $1\le x\le 2^k$ is odd. Note that if $x^3+3\equiv y^3+3\pmod{2^k}$, then $2^k\mid (x-y)(x^2+xy+y^2)$. Since $x,y$ are odd, the term $x^2+xy+y^2$ is odd, too. Hence, $2^k\mid x-y$. Since $1\le x<y\le 2^k$ odd, this is impossible. Thus, this polynomial traces all odd residues modulo $2^k$, in particular, there is an $x$ so that $x^3\equiv 2^k-3\pmod{2^k}$. Hence, we handled the powers of $2$. Blending Steps 2,3 into Step 1 we conclude.

Solution: Note $x^8-16$ works for all non-powers of 2 by Hensel's Lemma. $x^2+7$ deals with powers of 2.

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PS any $(x^2-k)(x^2+k)(x^2+1)$ for $k\equiv 1(\bmod\; 8), \sqrt{k}$ not integer should work. What a troll problem.