2.2. In the set of natural numbers, solve the equation

$$
7^{x}-32^{y}=1
$$
2.2. $(7-1)\left(7^{x-1}+7^{x-2}+\ldots+7+1\right)=62^{y-1}$. Follow $7^{x-1}+7^{x-2}+\ldots+7+1=2^{y-1}$.

If $y=1$, then $x=1$ is one solution.

Let $y>1$. Then $x$ is even and

$(7+1)\left(7^{x-2}+7^{x-4}+\ldots+7^{2}+1\right)=2^{y-1}$,

i.e. $\quad 7^{x-2}+7^{x-4}+\ldots+7^{2}+1=2^{y-4}$. For $y=4$ we get $x=2$ another solution.

For $y>4 \quad x$ must be divisible by 4. Then $\left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\ldots+7^{2}+1\right)=2^{y-4}$. The left side is divisible by 5, while the right side is not. The solutions are $x=1, y=1$ and $x=2, y=4$.