Let
$$x=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=2\frac{a^{4}+b^{4}}{a^{4}-b^{4}}.$$
Then
$$\frac{a^{4}+b^{4}}{a^{4}-b^{4}}+\frac{a^{4}-b^{4}}{a^{4}+b^{4}}=\frac{x}{2}+\frac{2}{x}.$$
We can repeat this to get
$$\frac{a^{8}+b^{8}}{a^{8}-b^{8}}+\frac{a^{8}-b^{8}}{a^{8}+b^{8}}=\frac{\frac{x}{2}+\frac{2}{x}}{2}+\frac{2}{\frac{x}{2}+\frac{2}{x}}=\frac{x}{4}+\frac{1}{x}+\frac{4x}{x^{2}+4}.$$
We have that $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}} \ (m \in \mathbb N)$$
$$\iff m - \sqrt{n + 2} = \sqrt{n + \sqrt{n + 2}} \iff (m - \sqrt{n + 2})^2 = n + \sqrt{n + 2}$$
$$\iff m^2 - (2m - \sqrt{n + 2})\sqrt{n + 2} = (\sqrt{n + 2} + 1)\sqrt{n + 2} - 2$$
$$\iff m^2 + 2 = (2m + 1)\sqrt{n + 2} \iff \sqrt{n + 2} = \frac{m^2 + 2}{2m + 1}$$
As an addition, $\dfrac{m^2 + 2}{2m + 1} \in \mathbb Q^+, \forall m \in \mathbb N \implies \sqrt{n + 2} \in \mathbb Q^+$
$\implies \sqrt{n + 2} \in \mathbb N \implies \dfrac{m^2 + 2}{2m + 1} \in \mathbb N \iff \dfrac{4(m^2 + 2) - (2m + 1)(2m - 1)}{2m + 1} \in \mathbb N$
$\iff \dfrac{9}{2m + 1} \in \mathbb N \iff 2m + 1\mid 9 \iff 2m + 1 \in \{1, 3, 9\} \iff m \in \{0, 1, 4\}$
We can set up a table for different value of $m$ and $\sqrt{n + 2}$.
$$\begin{matrix} m&& 0&& 1&& 4\\ \sqrt{n + 2} = \dfrac{m^2 + 2}{2m + 1}&& 2&& 1&& 2 \end{matrix}$$
$\iff n \in \{-1, 2\}$.
Plugging $n \in \{-1, 2\}$ in $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$, we have that $(m,n) = (1, -1)$ and $(m, n) = (4, 2)$ is the correct answer.
Best Answer
Write $$x^2(y-2)^2 = -y^3+3y^2-1$$
Since $y-2\mid -y^3+3y^2-1$ and $y\equiv 2\pmod{y-2}$ we have
$$0\equiv -y^3+3y^2-1 \equiv -8+12-1 \equiv 3 \pmod{y-2}$$ So $$y-2\mid 3\implies y-2\in\{1,-1,3,-3\}$$
so $$y\in\{3,1,5,-1\}$$ Checking each of them we are done.