The only such $n$ are $n \in \{1344,1728,1792,1920,1984\}.$ Let $p$ be an odd prime. Note that $p \mid x \iff p \mid y$. So, let $k$ = min($\nu_p(x), \nu_p(y)) >0$. Then notice that, $$2nk = \nu_p((x^2+y^2)^n) = \nu_p({(xy)}^{2016}) = 2016 (\nu_p(x) + \nu_p(y)) \geq 2016 \times 2k$$. So, $n \geq 2016.$ But, $$(xy)^{2016} = (x^2+y^2)^n \geq (2xy)^n > (xy)^n.$$ Hence, $n < 2016$ which is a contradiction. So no odd primes divide both $x$ and $y$. Furthermore, if $\nu_2(x) \neq \nu_2(y)$, we get a contradiction in similar fashion. So let $x = y = 2^m$. Thus our equation becomes, $$(2^{2m+1})^n = (2^{2m})^{2016}$$ implying, $n(2m+1) = 2m \times 2016$. This gives us that $2m +1 \mid 2016 = 2^5 \times 3^2 \times 7.$ Hence, $$2m+1 \in \{1, 3, 7, 9, 21, 63\} \implies m \in \{1,3,4,10,31\}$$giving us the possible values of $n \in \{1344,1728,1792,1920,1984\}.$