Find all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$.
If $y=1$, then $2 x \mid x^{2} \Leftrightarrow x=2 n, n \in \mathbb{N}$. So, the pairs $(x, y)=(2 n, 1), n \in \mathbb{N}$ satisfy the required divisibility.
Let $y>1$ such, that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. There exists $m \in \mathbb{N}$ such that

$$
x^{2}=m\left(2 x y^{2}-y^{3}+1\right), \text { i.e. } x^{2}-2 m y^{2} x+\left(m y^{3}-m\right)=0
$$

The discriminant of the last quadratic equation is equal to $\Delta=4 m^{2} y^{4}-4 m y^{3}+4 m$. Denote

$$
A=4 m\left(y^{2}-1\right)+(y-1)^{2}, \quad B=4 m\left(y^{2}+1\right)-(y+1)^{2}
$$

For $y>1, y \in \mathbb{N}$ and $m \in \mathbb{N}$ we have

$$
A>0, \quad B=4 m\left(y^{2}+1\right)-(y+1)^{2}>2\left(y^{2}+1\right)-(y+1)^{2}=(y-1)^{2} \geq 0 \Rightarrow B>0
$$

We obtain the following estimations for the discriminant $\Delta$ :

$$
\begin{aligned}
& \Delta+A=\left(2 m y^{2}-y+1\right)^{2} \geq 0 \Rightarrow \Delta\left(2 m y^{2}-y-1\right)^{2} .
\end{aligned}
$$

Because the discriminant $\Delta$ must be a perfect square, we obtain the equalities:

$$
\Delta=4 m^{2} y^{4}-4 m y^{3}+4 m=\left(2 m y^{2}-y\right)^{2} \Leftrightarrow y^{2}=4 m \Rightarrow y=2 k, k \in \mathbb{N}, m=k^{2}, k \in \mathbb{N}
$$

The equation $x^{2}-8 k^{4} x+k\left(8 k^{4}-k\right)=0$ has the solutions $x=k$ and $x=8 k^{4}-k$, where $k \in \mathbb{N}$.
Finally, we obtain that all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$, are equal to $(x, y) \in\left\{(2 k, 1),(k, 2 k),\left(8 k^{4}-k, 2 k\right) \mid k \in \mathbb{N}\right\}$.