Complexing

$\begin{aligned} z + \frac 1z & = \sqrt 3 & \small \color{#3D99F6} \text{Let }z = e^{ix} \\ e^{ix} + e^{-ix} & = \sqrt 3 \\ \frac 12 \left(e^{ix} + e^{-ix} \right) & = \frac {\sqrt 3}2 \\ \cos x & = \frac {\sqrt 3}2 \\ \implies x & = \frac \pi 6 \end{aligned}$

Then, we have:

$\begin{aligned} \frac {z^n+1}{z^n-1} & = \frac {e^{\frac {n \pi}6i}+1}{e^{\frac {n \pi}6i}-1} & \small \color{#3D99F6} \text{Multiply up and down by }e^{-\frac {n \pi}{12}i} \\ & = \frac {e^{\frac {n \pi}{12}i}+e^{-\frac {n \pi}{12}i}}{e^{\frac {n \pi}{12}i}-e^{-\frac {n \pi}{12}i}} \\ & = {\color{#3D99F6}i} \frac {\frac 12 \left(e^{\frac {n \pi}{12}i}+e^{-\frac {n \pi}{12}i}\right)}{-\frac {\color{#3D99F6}i}2 \left(e^{-\frac {n \pi}{12}i}-e^{\frac {n \pi}{12}i}\right)} \\ & = - i \frac {\cos \frac {n \pi}{12}}{\sin \frac {n \pi}{12}} \\ & = - i \cot \frac {n \pi}{12} \end{aligned}$

$\begin{aligned} \implies \frac {n \pi}{12} & = \frac {25 \pi}3 = \frac \pi 3 \\ \implies n & = \boxed{4} \end{aligned}$