Solve the following equation on the set of positive integers:

$$
x-y=\frac{x}{y}+\frac{x^{2}}{y^{2}}+\frac{x^{3}}{y^{3}}
$$
Let $x=x_{1} d$ and $y=y_{1} d$, where $d$ is the greatest common divisor of $x$ and $y$, and thus $x_{1}$ and $y_{1}$ are coprime. Substituting these expressions into the equation:

$$
d\left(x_{1}-y_{1}\right)=\frac{x_{1} y_{1}^{2}+x_{1}^{2} y_{1}+x_{1}^{3}}{y_{1}^{3}}
$$

Since the left side is an integer, the right side must also be an integer. For this to be true, the numerator of the right side, $x_{1}^{3}$, must be divisible by $y_{1}$. Since $x_{1}$ and $y_{1}$ are coprime, this is only possible if $y_{1}=1$. With this, the equation becomes:

$$
d\left(x_{1}-1\right)=x_{1}+x_{1}^{2}+x_{1}^{3}=\left(x_{1}-1\right)\left(x_{1}^{2}+2 x_{1}+3\right)+3
$$

For this to hold, it is necessary that $x_{1}-1$ is a divisor of 3. Since $x_{1}$ is a positive integer, this gives $x_{1}=2$, or $x_{1}=4$. Solving for $d$: $d=x_{1}^{2}+2 x_{1}+3+\frac{3}{x_{1}-1}$, which yields $d=14$, and $d=28$. Therefore, the solutions are: $x=28$, $y=14$, and $x=112$, $y=28$.