I have tried an elementary method to solve the equation $x^{2}-7^{y}=2$ and have given up this ideal at last. Then I try to solve it in $Q(\sqrt{2})$. Since $Q(\sqrt{2})$ has infinity many units $\pm(\pm 1+\sqrt{2})^{n}$, $n=0,1,2,\cdot\cdot\cdot$, the proof would become more difficult. Note that $(x,y)=(\pm 3,1)$ are integer solutions.
Update: An elementary number theory method on the equation which has the only integer solutions $(x,y)=(\pm 3,1)$ can be found in the direction of Pell equation and I have finished it many days before. This method has been mentioned in Mikael Jensen's comments below but there still needs much more other skills.
Now, my question is:
Does there exist a method of algebraic number theory to solve the equation?
Just some comments. There are primitive solutions to $$ x^2 - 2 z^2 = 7^y $$ for every positive $y.$ Indeed, if $z_n \equiv 5 x_n \pmod 7$ and $$ x_n^2 - 2 z_n^2 = 7^n, $$ we may take $$ x_{n+1} = 3 x_n + 2 z_n, \; \; z_{n+1} = x_n + 3 z_n $$ to get a primitive solution to $$ x_{n+1}^2 - 2 z_{n+1}^2 = 7^{n+1}, \; \; \; z_{n+1} \equiv 5 x_{n+1} \pmod 7. $$
Meanwhile, given any such $(x,z)$ pair, we get the same $7^n$ with either $$ (3 x + 4 z, 2 x + 3 z) $$ or $$ (3 x - 4 z, -2 x + 3 z). $$ After taking absolute values, you can get some pretty small values of $z$ this way: $$ 9^2 - 2 \cdot 4^2 = 49, $$ $$ 19^2 - 2 \cdot 3^2 = 343, $$ $$ 51^2 - 2 \cdot 10^2 = 2401, $$ and so on.
I will need to think about whether all solutions of $x^2 - 2 z^2 = 7^y$ arise this way. If so, it is a start.