Answer: $\pm x^k$ for $k\in\mathbb Z$ and $0\le k\le n-1$. All work by letting $m=2n$. Let $Q(x)=\sum_{i=0}^{n-1}a_ix^i$. Let $\omega$ be a primitive $n$-th root of unity. It follows that $Q(\omega^i)$ is an $m$-th root of unity. Hence $$n=\sum_{i=0}^{n-1}|Q(\omega^i)|^2=\sum_{i=0}^{n-1}Q(\omega^i)Q(\omega^{-i})=n\sum_{i=0}^{n-1}a_i^2.$$ Hence $\sum_{i=0}^{n-1}a_i^2=1$. As $a_i$ are all integers, it follows that $a_i$ are all zero, except one of them which is $\pm 1$.

Another proof: Let $Q(x)=\sum_{i=0}^{n-1}a_ix^i$. Let $\omega$ be a primitive $n$-th root of unity. The zero polynomial does not work so there exists an integer $p\in[0,n-1]$ such that $a_p\ne 0$. Then $|a_p|\ge 1$ since $a_p$ is an integer. By basic properties of Fourier series, we know that $$a_p=\frac{1}{n}\sum_{j=0}^{n-1}Q(\omega^j)\omega^{-pj}.$$The conditions imply that $Q(\omega^j)$ is an $m$-th root of unity for all $j=0,\ldots,n-1$. Hence by triangle inequality, $$1\le |a_p|=\left|\frac{1}{n}\sum_{j=0}^{n-1}Q(\omega^j)\omega^{-pj}\right|\le\frac{1}{n}\sum_{j=0}^{n-1}\left|Q(\omega^j)\omega^{-pj}\right|=1.$$Since equality holds, there exists a complex $c$ with $|c|=1$ and $Q(\omega^j)=c\omega^{pj}$ for all $j=0,\ldots,n-1$. Since $\deg(Q)<n$, it follows that $Q(x)\equiv cx^p$ as a polynomial in $\mathbb C[x]$. Hence $c$ is an integer and $|c|=1$, so $c\in\{-1,1\}$.

Very similar proof to math90's first proof, but just a dumbed down version for amateurs like me: Same as math90's post, we will let $w$ be the primitive $n$th root of unity, and note that $Q(w^i)$ is an $m$th root of unity for all $1\leq i \leq n$. Let $a_i$ be the $i$th coefficient of $Q(x)$, with $a_0$ being the constant term, and let $k$ be the degree of $Q$. We will assume $Q$ has a constant term. $\sum^{n-1}_{i=0} Q(w^i) = \sum^{k-1}_{i=0}(\sum^{n-1}_{j=0}a_i(w^j)^i) = na_0$ (since $\sum^{n-1}_{j=0}a_i(w^j)^i=0$ for all $i$ except for 0) However, notice that (by the triangle inequality) $|\sum^{n-1}_{i=0} Q(w^i)| \leq n$ but we also have $|na_0| \geq n$, and the only way for both to be equal is if $Q(w^i)$ = 1 for all $i \leq n$, but this means $Q(x) - 1$ has $n$ roots which is not allowed, thus we have a contradiction. Therefore $Q(x)$ does not have a constant term, thus we can write $Q(x) = R(x)x^l$ where $R(x)$ has integer coefficients and also has a constant term. Now we can repeat the same argument for $R$ since we have that $R(w^i)$ is an $m$th root of unity. We continue doing this until we get that $Q(x) = \pm x^k$ for any integer $k$.