System equavalent to $ (x+u\sqrt 2 )(y-v\sqrt 2 )=-16-12\sqrt 2$. $ 16^2+2*12^2=2^5*17$. There are only one way factorizate $ 17=(5+2\sqrt 2)(5-\sqrt 2)$. Uniques are $ \pm 1, \pm 1\pm \sqrt 2$ and 2 ways factorizate $ 4=2*2=(2+\sqrt 2)(2-\sqrt 2 )$. It give all ways factorizate $ -16-12\sqrt 2$ or all solutions these system.

Rust wrote: System equavalent to $ (x + u\sqrt 2 )(y - v\sqrt 2 ) = - 16 - 12\sqrt 2$. $ 16^2 + 2*12^2 = 2^5*17$. There are only one way factorizate $ 17 = (5 + 2\sqrt 2)(5 - \sqrt 2)$. Uniques are $ \pm 1, \pm 1\pm \sqrt 2$ and 2 ways factorizate $ 4 = 2*2 = (2 + \sqrt 2)(2 - \sqrt 2 )$. It give all ways factorizate $ - 16 - 12\sqrt 2$ or all solutions these system. I don't think that the above proof is correct. In the UFD $ \mathbb Z[\sqrt2]$, the absolute norm of $ - 16 - 12\sqrt2$ is $ |16^2 - 2\cdot 12^2| = 2^5$, (not $ 16^2 + 2\cdot 12^2 = 17 \cdot 2^5$), and there are infinite number of units, viz. $ \pm(1 + \sqrt2)^n$, $ n\in\mathbb Z$, (not just $ \pm 1$, $ \pm 1\pm \sqrt 2$).

You are right. Solutions are $ x + y\sqrt 2 = \pm (\sqrt 2)^m(1 + \sqrt 2)^k, y - v\sqrt 2 = \mp (\sqrt 2 - 1)^{k - 2}(\sqrt 2)^{5 - m},m = 0,1,2,3,4,5, k\in Z.$