$12 \cdot 91$ Try to find all natural number solutions of the equation $7^{x}-3 \cdot 2^{y}=1$.
(16th All-Russian Mathematical Olympiad, 1990)
Given the equation can be transformed into
$$
\frac{7^{x}-1}{7-1}=2^{y-1} .
$$

This is equivalent to $\quad 7^{x-1}+7^{x-2}+\cdots+1=2^{y-1}$.
If $y=1$, then $x=1$, thus we obtain the first set of natural number solutions $(x, y)=$
$(1,1)$.
Assume $y>1$, in this case, the right side of equation (1) is even, and the left side is the sum of $x$ odd numbers, hence $x$ must be even.
Thus, equation (1) can be rewritten as
$$
\begin{array}{c}
7^{x-2}(7+1)+7^{x-4}(7+1)+\cdots+(7+1)=2^{y-1}, \\
(7+1)\left(7^{x-2}+7^{x-4}+\cdots+1\right)=2^{y-1}, \\
7^{x-2}+7^{x-4}+\cdots+1=2^{y-4} .
\end{array}
$$

From this, we know that
$$
y \geqslant 4 \text {. }
$$

If $y=4$, then $x=2$, thus we obtain the second set of natural number solutions $(x, y)=$ $(2,4)$.

If $y>4$, then the right side of equation (2) is even, and the left side is the sum of $\frac{x}{2}$ odd numbers, therefore, $x$ must be a multiple of 4.
Thus, equation (2) can be rewritten as
$$
\begin{array}{c}
7^{x-4}\left(7^{2}+1\right)+7^{x-8}\left(7^{2}+1\right)+\cdots+\left(7^{2}+1\right)=2^{y-4}, \\
\left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\cdots+1\right)=2^{y-4},
\end{array}
$$

From (3), we see that $2^{y-4}$ must be divisible by $7^{2}+1=50$, which is impossible.
Therefore, when $y>4$, the equation has no natural number solutions.
In summary, the equation has only two sets of natural number solutions:
$$
(x, y)=(1,1),(2,4) .
$$