The number of prime pairs of $x^2-2y^2=1$

Question

How to find the number of pairs of positive integers $(x,y)$ where $x$ and $y$ are prime numbers and $x^2-2y^2=1$?

I am not getting any clue here.

Answer 1

Flip it around:

$$(x-1)(x+1)=2y^2$$

Because of factor $2$, left hand side must be divisible by $2$. But if one of the factors is divisible by $2$, both are, so overall it's divisible by $4$. Draw your own conclusions from here.

Answer 2

Since $2y^2$ is even, $x$ must be odd so that $x^2$ can be odd as well. This much should be obvious.

From there it's not too much of a leap to find that $x^2 \equiv 1 \pmod 4$. If $y$ is odd as well, then $y^2 \equiv 1 \pmod 4$, too, but $2y^2 \equiv 2 \pmod 4$, which means that $x^2 - 2y^2 \equiv 3 \pmod 4$, but clearly $1 \equiv 1$, not $3 \pmod 4$.

Therefore $y$ must be even, so that $2y^2 \equiv 0 \pmod 4$, and consequently $x^2 - 2y^2$ can be congruent to $1 \pmod 4$ as is necessary to solve the equation.

Of course 2 is prime, and so is $-2$. If $y = \pm 2$, then $x = \pm 3$. And if $y = 0$, then $x = 1$. The only remaining possibilities are composite even numbers.