Solution:

It is easy to check that $\left( \pm 2^{r}, 0, 2r \right)$ and $\left(0, \pm 2^{r}, 2r+1 \right)$ satisfy this equation for any nonnegative integer $r$. We will show that these are all the solutions by an infinite descent method.

So suppose we have some solution $\left(x_{0}, y_{0}, n_{0}\right)$. If $x_{0}$ is odd, then $2^{n_{0}}$ is odd, which forces $n_{0}=0$ and then $x_{0}^{2} + 2 y_{0}^{2} = 1$, so $y_{0} = 0$ (or else $2 y_{0}^{2} > 1$ already) and then $x_{0} = \pm 1$.

On the other hand, if $x_{0}$ is even, we can let $x_{0} = 2 x_{0}'$ and then $4 x_{0}'^{2} + 2 y_{0}^{2} = 2^{n_{0}} \Rightarrow y_{0}^{2} + 2 x_{0}'^{2} = 2^{n_{0}-1}$, so $\left(x_{1}, y_{1}, n_{1}\right) = \left(y_{0}, x_{0}/2, n_{0}-1\right)$ is another solution to our equation, where $n_{0}$ has been replaced by $n_{0}-1$. Now if $x_{1}$ is even, we can repeat this construction to get another new solution $\left(x_{2}, y_{2}, n_{2}\right)$ with $n_{0}-1$ replaced by $n_{0}-2$, and so on. These integers $n$ cannot go on decreasing forever, since there does not exist an integral solution where $n < 0$. Thus, eventually our process terminates, which means we get to a solution $\left(x_{k}, y_{k}, n_{k}\right)$ with $x_{k}$ odd. By the above, this is possible only if $x_{k} = \pm 1, y_{k} = 0, n_{k} = 0$.

On the other hand, the above construction can be performed in reverse: we have $x_{i} = 2 y_{i+1}, y_{i} = x_{i+1}, n_{i} = n_{i+1} + 1$ for each value of $i \geq 0$. Now we claim that $\left(x_{i}, y_{i}, n_{i}\right) = \left( \pm 2^{(k-i)/2}, 0, k-i \right)$ when $k-i$ is even, and $\left(0, \pm 2^{(k-i-1)/2}, k-i \right)$ when $k-i$ is odd. The proof is by downward induction: the base case $i = k$ is certainly true; given that the statement holds for some $i > 0$, it is simple algebra to check that it holds for $i-1$ by applying our reverse construction. Thus, the claim is true for each $i \geq 0$. In particular, $\left(x_{0}, y_{0}, n_{0}\right) = \left( \pm 2^{k/2}, 0, k \right)$ or $\left(0, \pm 2^{(k-1)/2}, k \right)$, which fits the form above. So, every solution is of this form.