Example 3 Find the positive integer solutions of the equation $7^{x}-3 \times 2^{y}=1$.
(16th All-Russian Mathematics Competition)
When $1 \leqslant y \leqslant 4$, the equation has two sets of solutions: $x=y=1$ and $x=2, y=4$.
When $y \geqslant 5$, by reversing the geometric series sum formula, we have
$$
\begin{aligned}
\frac{7^{x}-1}{7-1} & =7^{x-1}+7^{x-2}+\cdots+1 \\
& =\frac{3 \times 2^{y}}{7-1}=2^{y-1} .
\end{aligned}
$$

Therefore, $x$ must be even. Hence, we can set $x=2 m(m \in \mathbf{N})$. Reversing the geometric series sum formula, we have
$$
\begin{aligned}
\frac{7^{2 m}-1}{7^{2}-1} & =7^{2 m-2}+7^{2 m-4}+\cdots+1 \\
& =\frac{3 \times 2^{y}}{7^{2}-1}=2^{y-4} .
\end{aligned}
$$

Therefore, $m$ must be even. We can set $x=4 n(n \in \mathbf{N})$, and reversing the geometric series sum formula, we have
$$
\begin{aligned}
\frac{7^{4 n}-1}{7^{4}-1} & =7^{4 n-4}+7^{4 n-8}+\cdots+1 \\
& =\frac{3 \times 2^{y}}{7^{4}-1}=\frac{3 \times 2^{y}}{48 \times 50} .
\end{aligned}
$$

Clearly, $3 \times 2^{y}$ cannot be divisible by 5, so when $y \geqslant 5$, the equation has no positive integer solutions.

Therefore, the original equation has only two sets of solutions: $x=y=1$ and $x=2, y=4$.