Solve the following equation in the set of integers:

$$
2^{\frac{x-y}{y}}-\frac{3}{2} y=1
$$
Solution. Clearly $y \neq 0$. Let's transform the equation:

$$
1=2^{\frac{x-y}{y}}-\frac{3}{2} y=2^{\frac{x}{y}-1}-\frac{3}{2} y=\frac{2^{\frac{x}{y}}}{2}-\frac{3}{2} y
$$

From this,

$$
2^{\frac{x}{y}}-2=3 y
$$

The number on the right side is divisible by 3. The expression on the left side

$$
2^{\frac{x}{y}}-2>-2
$$

because powers of 2 are positive numbers and (1) implies that $2^{\frac{x}{y}}$ must be an integer. Therefore, $\frac{x}{y}$ is also an integer. From (1) and (2), it follows that $3 y>-2$, and since $y \neq 0$, it follows that $y>0$.

Let's examine equation (1). On the left side, there is a number that is 2 less than a power of 2, which is divisible by 3.

Case I: If $\frac{x}{y}=2 n$, then $2^{\frac{x}{y}}=2^{2 n}=\left(2^{n}\right)^{2}$, which gives a remainder of 1 when divided by 3, and thus the remainder of the left side when divided by 3 is 2. Therefore, there is no solution in this case.

Case II: If $\frac{x}{y}=2 n+1$, then $2^{\frac{x}{y}}=2^{2 n+1}=2 \cdot 2^{2 n}$, which gives a remainder of 2 when divided by 3. Thus, the left side is divisible by 3, and in this case, there is a solution: $2^{2 n+1}-2=3 y$, from which

$$
y=\frac{2^{2 n+1}-2}{3}
$$

Since $\frac{x}{y}=2 n+1$,

$$
\frac{x}{\frac{2^{2 n+1}-2}{3}}=2 n+1, \quad \text { hence } \quad x=\frac{(2 n+1)\left(2^{2 n+1}-2\right)}{3}
$$

The solution is:

$$
x=\frac{(2 n+1)\left(2^{2 n+1}-2\right)}{3}, \quad y=\frac{2^{2 n+1}-2}{3}, \quad \text { where } n \in \mathbb{Z}^{+}
$$