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The answer is $\boxed{\binom{m}{n}\binom{n}{m-n}}$. Let $w$ be the sixth root of unity. Notice all the $z_i$ must be $1$, $w$, or $w^5$. This means $z_i^7=z^i$. In order for the sum to have no imaginary part we must have the same amount of $z_i$ being $w$ and $w^5$, call this amount $a$. Then the sum of the $z_i$ evaluates to $m-a$ so we must have $a=m-n$ leading to the claimed solution.

For each $k\in\{1,2,\dots,m\}$ consider \begin{align*} z_k^3-2z_k^2+2z_k-1=0&\Rightarrow (z_k^3-1)-2z_k(z_k-1)=0\\ &\Rightarrow (z_k-1)(z_k^2+z_k+1)-2z_k(z_k-1)=0\\ &\Rightarrow (z_k-1)(z_k^2-z_k+1)=0 \end{align*}yielding $z_k\in\{1,\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right),\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\}$. For convenience we define $a=\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)$ and $b=\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)$. By De Moivre's theorem, $z_k^7\in\{1,\cos\left(\frac{7\pi}{3}\right)+i\sin\left(\frac{7\pi}{3}\right),\cos\left(-\frac{7\pi}{3}\right)+i\sin\left(-\frac{7\pi}{3}\right)\}$ and note that \begin{align*} \cos\left(\frac{7\pi}{3}\right)+i\sin\left(\frac{7\pi}{3}\right)&=\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)=a\\ \cos\left(-\frac{7\pi}{3}\right)+i\sin\left(-\frac{7\pi}{3}\right)&=\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)=b. \end{align*}Now, $z_k^7\in\{1,a,b\}$ for every $k=1,2,\dots,m$. Suppose that $s,t$ be integers such that $$s=\#\{z_k\mid k\in\{1,2,\dots,m\},z_k^7=a\},\;\; t=\#\{z_k\mid k\in\{1,2,\dots,m\},z_k^7=b\}.$$Since $z_1^7+z_2^7+\cdots+z_m^7=n$ implies that $Im(z_1^7+z_2^7+\cdots +z_m^7)=0$ which is $$Im\bigl(1\cdot (m-s-t)+a\cdot s+b\cdot t\bigr)=0\Rightarrow Im(as+bt)=0.$$It is equivalent to $Im(i\sin\left(\frac{\pi}{3}\right)\cdot s-i\sin\left(\frac{\pi}{3}\right)\cdot t)=0\Rightarrow \sin\left(\frac{\pi}{3}\right)\cdot s=\sin\left(\frac{\pi}{3}\right)\cdot t$, so $s=t$. Plugging back into the equation $z_1^7+z_2^7+\cdots+z_m^7=n$, we get that $$1\cdot (m-2s)+\frac{s}{2}+\frac{s}{2}=n\Rightarrow s=m-n$$Since $z_1,z_2,\dots,z_k$ is the permutation of $(m-n)$ $a$'s, $(m-n)$ $b$'s and $(2n-m)$ $1$'s, therefore the number of solutions of the given equation is exactly $$\frac{m!}{(m-n)!(m-n)!(2n-m)!}.$$