Find all pairs of integers $(x, y)$ such that

$$
x^{4}-2 y^{2}=1
$$
The equation is written as

$$
2 y^{2}=x^{4}-1=(x-1)(x+1)\left(x^{2}+1\right)
$$

Since the left-hand side is even, $x$ is odd and the right-hand side is divisible by 8, and $y$ is even. We replace $x$ by $2 x^{\prime}+1$ and $y$ by $2 y^{\prime}$. The equation becomes

$$
y^{\prime 2}=x^{\prime}\left(x^{\prime}+1\right)\left(2 x^{\prime 2}+2 x^{\prime}+1\right)
$$

It is easy to see that these three terms are pairwise coprime (we form a combination that gives 1). Let $p$ be an odd prime factor of $y^{\prime}$, and $k$ its valuation. So $p^{2 k} \mid x^{\prime}\left(x^{\prime}+1\right)\left(2 x^{\prime 2}+2 x^{\prime}+1\right)$, and since all these terms are coprime, this means that $p^{2 k}$ divides one of the three terms. This means that the three terms are perfect squares. But the only squares of the form $x$ and $x+1$ are 0 and 1, so the only solution is $x=1$ and $y=0$.