Here is a sketch of an argument:

Since $\tan(z)= \frac{1}{i} \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$, it is close to $\frac{1}{i}$ for $\Im(z)\gg 0$ and close to $-\frac{1}{i}$ for  $\Im(z)\ll 0$. In particular, $|\tan(z)|$ is bounded above for large values of $\Im(z)$ and $z^n$ is bounded below for large values of $\Im(z)$, so $\tan(z)-z^n$ has no zeros outside of the region $-C\leq \Im(z)\leq C$.

We now let $\gamma$ be the contour given by the boundary of the rectangle $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$, $-C\leq \Im(z)\leq C$, minus two small disks of radius $\varepsilon$ around $\pm \frac{\pi}{2}$.

$|\tan(z)|$ stays bounded on $\gamma$, say by some $B$. For large enough $n$, we have that $|z^n|$ is strictly bigger than $B$ on all of $\gamma$, since $\gamma$ contains no $z$ with $|z|\leq 1$. So we can apply Rouche's theorem to see that $\tan(z)-z^n$ and $z^n$ have the same number of zeroes in our region, which is $n$.

This proves that in the region $-\frac{\pi}{2} \leq \Re(z)\leq \frac{\pi}{2}$ minus two small circles around $\pm \frac{\pi}{2}$, $\tan(z)-z^n$ has indeed exactly $n$ zeroes for large enough $n$ (where the precise numbers depend on the $\varepsilon$ you chose). However, there are also one or two zeroes in those circles: As $\tan(z)-z^n$ is negative for something like $z=1.5$, but $\tan(z)$ goes to $+\infty$ as $z\to \frac{\pi}{2}$ from below in the reals, it must become positive again. So there is at least one more zero there, and for odd $n$ another one near $-\frac{\pi}{2}$ by symmetry. I believe these are the only ones by looking at the fact that $\tan(z)\approx -\frac{1}{z-\frac{\pi}{2}}$, and one can probably see this rigorously by similar tricks as above.