18.4.13 *Find all natural number solutions to the equation: $7^{x}-3 \cdot 2^{y}=1$.


The original equation is equivalent to $\frac{7^{x}-1}{7-1}=2^{y-1}$, which is also $7^{x-1}+7^{x-2}+\cdots+1-2^{y-1}$.
If $y=1$, then $x=1$.
Assume $y>1$, then $7^{x} \equiv(-1)^{x} \equiv 1(\bmod 4)$, so $x$ is an even number $2u$.
$$
7^{2u}-1=\left(7^{u}+1\right)\left(7^{u}-1\right)=3 \times 2^{y} \text{. }
$$

And $\left(7^{u}+1,7^{u}-1\right)=\left(7^{u}+1,2\right)=2$, so from (1) we get,

The first equation set obviously has no solution $\left(7^{u} \neq 3\right)$. The second equation set gives $u=1, x=2, y=4$.
The solutions to this problem are $(x, y)=(1,1),(2,4)$.