Two variable Equation? Easy enough!

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

If $x+1=2 \Rightarrow x=1$ which is not allowed.

If $x+1=y \ , \ x-1=2y \Rightarrow \ x \ is \ odd \ \Rightarrow x=1 \ , \ y=2$ which is not allowed.

If $x+1=2y \ , \ x-1=y \Rightarrow \ x \ is \ odd \ \Rightarrow x=3 \ , \ y=2$ .which is absolutely possible.

If $x-1=2 \Rightarrow x=3 \ , \ y=2$ which is absolutely possible.

Thus , $(x,y) = (3,2)$ is the only possible solution, and required answer $=2+3=5$ .

$x^2=2y^2+1$ gives that x must be an odd integer.

Let x= $2n+1$ This implies ${{(2n+1)}}^2=4n^2+4n+1=2y^2+1$

$y^2=2n(n+1)$ Which implies $y^2$ is even, So, y has to be even.

Hence, y=2 (Only even prime number). Which gives $x=3$

$x+y=5$

No other values of $x,y$ satisfy the equation above, as y has to be even, and 2 is the only even prime number.

Answer:- $\boxed {5}$