Solve $7^{x}-3 \cdot 2^{y}=1$
We are looking to factorize: $7^{x}-1=3 \cdot 2^{y}$.

$$
(7-1)\left(7^{x-1}+7^{x-2}+\cdots+7+1\right)=3 \cdot 2^{y}
$$

This gives

$$
7^{x-1}+7^{x-2}+\cdots+7+1=2^{y-1}
$$

This provides a first solution: $y=1$ and $x=1$. (We can verify $7-3 \cdot 2=1$)

Next, we notice that the term on the left is a sum of $x$ odd terms, so $x$ must be even. We can then group the terms in pairs.

$$
\begin{aligned}
2^{y-1} & =\left(7^{x-1}+7^{x-2}\right)+\left(7^{x-1}+7^{x-2}\right)+\cdots+(7+1) \\
& =(7+1) 7^{x-2}+(7+1) 7^{x-4}+\cdots+(7+1) \\
& =(7+1)\left(7^{x-2}+7^{x-4}+\cdots+7^{2}+1\right)
\end{aligned}
$$

Thus,

$$
7^{x-2}+7^{x-4}+\cdots+7^{2}+1=2^{y-4}
$$

This gives a second solution: $y=4$ and $x=2$. We can verify $49-3 \cdot 2^{4}=1$.

We can continue in the same manner: there must be an even number of terms in the sum on the left.

$$
\begin{aligned}
2^{y-4} & =\left(7^{x-2}+7^{x-4}\right)+\left(7^{x-6}+7^{x-8}\right)+\cdots+\left(7^{2}+1\right) \\
& =\left(7^{2}+1\right) 7^{x-4}+\left(7^{2}+1\right) 7^{x-8}+\cdots+\left(7^{2}+1\right) . \\
& =\left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\cdots+7^{2}+1\right)
\end{aligned}
$$

But this is not possible because $7^{2}+1$ does not divide $2^{y-4}$. The only two solutions are therefore $x=1, y=2$ and $x=2, y=4$.