Given $z \in \mathbb{C}$ with $|z| > 1$, choose any $r > 1$. We have
$|1 + z^{2n}| \geq \left|1 - |z|^{2n}\right| > \frac{|z|}{r}^{2n}$ for large enough $n$. The first inequality comes from $|a + b| \geq ||a| - |b||$, which is always true, and the second one comes from the limit $\lim_{n \rightarrow \infty} \frac{\left|1 - |z|^{2n}\right|}{|z|^{2n}} = 1,$ and the choice of $r$, which makes $1/r$ less than $1$. Now, applying the root test as azarel suggested, we get $\limsup_n \sqrt[n]{\left|\frac{z^n}{1 + z^{2n}}\right|} < \limsup_n \frac{|z|}{\frac{|z|^2}{\sqrt[n]{r}}} = \frac{1}{|z|} < 1.$ Therefore the series converges for $|z| > 1$.
As Greg Martin noted, this series may be written as $\sum_{n = 0}^{\infty} \frac{1}{z^n + z^{-n}}$ which shows that its behavior must be the same for $|z| > 1$ or $|z| < 1$, so that we get convergence on $\mathbb{C}\backslash\mathbb{S}^1$. The behavior for $|z| = 1$ seems more complicated, though, so I'll leave to someone more experienced to answer. Hope this helps!