5. It is obvious that $x=y=0$ are roots of this equation. Let now $x, y>0$ be natural numbers greater than or equal to 1. Write the equation in the form: $(\sqrt{2 x}-1)(\sqrt{2 y}-1)=1$. Let for definiteness $y \geq x$, then $\sqrt{2 x}-1 \leq 1$, i.e. $2 x \leq 4, x \leq 2 \Rightarrow x=1, x=2$. For $x=1$ we get: $(\sqrt{2}-1)(\sqrt{2 y}-1)=1 \Rightarrow y=3+\sqrt{2}$ - a non-integer number. For $x=2$ we get $y=2$, hence $x=y=2$ is the only solution that is strictly greater than zero.
Answer: $x=y=0, x=y=2$.