Find natural number solutions to $x^2-2y^2=2$ One solution I found is $x=2, y=1$. But my question is: 
Is this the only natural solution?

With an equation of the form $x^2-Dy^2=N$, whenever $(x, y)$ is a solution, $(ax+by, cx+dy)$ is a solution for some $a, b, c, d$ which depend on the coefficients of the original equation but not on $x$ or $y$. Here's one way to find those coefficients $a, b, c, d$ in the OP's case:
\begin{align*}
(ax+by)^2-2(cx+dy)^2 &= 2\\
\implies (a^2-2c^2)x^2+2(ab-2cd)xy+(b^2-2d^2)y^2 &= 2\\
\implies a^2-2c^2 &= 1;\\
ab &= 2cd;\\
b^2-2d^2 &= -2.
\end{align*}
So if you can find a solution to $a^2-2c^2 = 1$, you may take $d=a$ and $b=2c$. $a=3, b=2$ works, so we get the recurrence which Robert Israel pointed out.
More generally, if the original equation is $x^2-Dy^2=N$,
\begin{align*}
(ax+by)^2-D(cx+dy)^2 &= N\\
\implies (a^2-Dc^2)x^2+2(ab-Dcd)xy+(b^2-Dd^2)y^2 &= N\\
\implies a^2-Dc^2 &= 1;\\
ab &= Dcd;\\
b^2-Dd^2 &= -D\\
\implies d^2-\frac{b^2}{D} &= 1\\
\implies d^2-D\Big(\frac{b}{D}\Big)^2&=1
\end{align*}
So again, if you can find a solution to $a^2-2c^2 = 1$, you may take $d=a$ and $b=Dc$.
