21. When $x=1$, it is known that $y=1$. Now consider the case $x \geqslant 2$, at this time $y \geqslant 4$, taking both sides modulo 8, we should have
$$(-1)^{x} \equiv 1(\bmod 8)$$

Thus $x$ is even, let $x=2 m$, then
$$\left(7^{m}-1\right)\left(7^{m}+1\right)=3 \cdot 2^{y}$$

Since $7^{m}-1$ and $7^{m}+1$ differ by 2, exactly one of them is a multiple of 3, there are two scenarios: $\square$
We get
Thus
$$\begin{array}{c}
7^{m}-1=3 \cdot 2^{u} \\
u=1 \\
y=u+v=4
\end{array}$$

If $m>1$, then
$$7^{m}+1=8 \cdot\left(7^{m-1}-7^{m-2}+\cdots-7+1\right)$$

Here, $7^{m-1}-7^{m-2}+\cdots-7+1$ is the sum of an odd number of odd numbers, which contradicts $7^{m}+1=2^{u}$.
In summary, the pairs $(x, y)$ that satisfy the conditions are $(1,1),(2,4)$.
21 Find all positive integer solutions to the indeterminate equation $7^{x}-3 \cdot 2^{y}=1$.

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