8.16 The number of positive integer pairs $(x, y)$ that satisfy the equation $7^{x}-3 \cdot 2^{y}=1$ is
(A) 0.
(B) 1.
(C) 2.
(D) Infinitely many.
(High School Mathematics Competition in Fuzhou City, Fujian Province, China, 1993)
[Solution] The given equation can be transformed into
$\frac{7^{x}-1}{7-1}=2^{y-1}$, which is $7^{x-1}+7^{x-2}+\cdots+1=2^{y-1}$.
If $y=1$, then $x=1$, i.e., $(1,1)$ satisfies the equation.
If $y>1$, since the right side of (1) is even, the left side must be an even number of odd numbers, which means $x$ is even, so the equation can be rewritten as

i.e., $\square$
$$
\begin{array}{c}
(7+1)\left(7^{x-2}+7^{x-4}+\cdots+1\right)=2^{y-1}, \\
7^{x-2}+7^{x-4}+\cdots+1=2^{y-4} .
\end{array}
$$

Then $y \geqslant 4$. We have $x=2, y=4$.
When $y>4$, the right side of equation (2) is even, and the left side is the sum of $\frac{x}{2}$ odd numbers, so $4 \mid x$. Thus, equation (2) can be rewritten as
$$
\left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\cdots+1\right)=2^{y-4},
$$
Since $7^{2}+1=50$, then $50 \mid 2^{y-4}$, which is impossible.
Therefore, the equation has no solution when $y>4$.
In summary, the original equation has only two solutions: $(1,1)$ and $(2,4)$.
Thus, the answer is $(C)$.