Caseworking!

Rearranging the equation yields $y=\sqrt{\dfrac{x^2 - 2}{5}}$ . This can be an integer only if $x^2 - 2$ is divisible by $5$ . We consider the quadratic residues moduluo $5$ : $\begin{aligned} 0^2 &\equiv 0 \mod 5 \\ 1^2 &\equiv 1 \mod 5 \\ 2^2 &\equiv 4 \mod 5 \\ 3^2 &\equiv 4 \mod 5 \\ 4^2 &\equiv 1 \mod 5 \end{aligned}$

Thus, no square is congruent to $2\bmod 5$ and, therefore, $x^2-2$ cannot be divisible by $5$ for any integer $x$ . So there is no integer $x$ for which $y$ is also an integer and, therefore, there are $\boxed{0}$ ordered pairs $(x,y)$ satisfying the given equation.

$x=\sqrt {2+5y^2}$

No matter what integer $y$ is, $5y^2$ will always end in either 0 or 5. So $2+5y^2$ will always end in either 2 or 7.

No integer has its square's unit place as 2 or 7.

Thus, no such pair exists.

Using the same condition as @Jordan Cahn ,

$y = \sqrt{\dfrac {x^2 - 2}{5}}$

$(x^2 - 2)$ can only be divisible by $5$ when the ones digit of $x^2$ will be either 2 or 7.

Note that there are no integers which when squared will give you 2 or 7 as their ones digits.

The only digits at ones place that we get after squaring integers are 0 , 1 , 4 ,9 , 6 and 5.

Hence, $\boxed{NO}$ integer solutions.