Answer: 71

Small Hint:

Substituting one equation into the other gives $z^{143} = z,$ so $z$ is a $142$nd root of unity

Big Hint:

Write $z = \cos\frac{2k\pi}{142} + i\sin\frac{2k\pi}{142}$ and reduce the fraction in the angle to lowest terms

Solution:

Because $0 \lt m \lt n,$ the specified imaginary part is positive, so $z \ne 0.$ Substituting, $z = w^{11} = (z^{13})^{11} = z^{143},$ and hence $z^{142} = 1.$ Conversely, any $142$nd root of unity $z$ works with $w = z^{13},$ since then $w^{11} = z^{143} = z.$

Hence $z = \cos\frac{2k\pi}{142} + i\sin\frac{2k\pi}{142}$ for some integer $k,$ and the imaginary part of $z$ is $\sin\frac{k\pi}{71}.$ The sine specified in the problem is positive, so we may take $1 \le k \le 70.$ Since $71$ is prime, $\frac{k}{71}$ is already in lowest terms, matching the required form $\sin\left(\frac{m\pi}{n}\right)$ with $m \lt n.$ Thus $n = 71.$