Three, find all positive integer solutions $(x, y, z)$ of the indeterminate equation $1+2^{x} \times 7^{y}=z^{2}$.
(Weng Shiyou supplied the problem)
Three, the original equation can be transformed into
$$
2^{x} \times 7^{y}=z^{2}-1=(z+1)(z-1) \text {. }
$$

It is easy to see that $z$ is an odd number, and $z+1$ and $z-1$ are two consecutive even numbers, with only one of them being a multiple of 4.

Since $(z+1)-(z-1)=2$, they cannot both be multiples of 7.
From the first equation, we get $2=2 \times 7^{y}-2^{x-1}$.
Clearly, $x \neq 1,2$.
When $x \geqslant 3$, $7^{y}-2^{x-2}=1$, taking both sides modulo 7 gives
$$
2^{x-2}=6(\bmod 7) \text {. }
$$

However, $2^{k} \equiv 1,2,4(\bmod 7)$, so this equation does not hold.

From the second equation, we get $2=2^{x-1}-2 \times 7^{y}$, it is easy to see that $x=5, y=1$ is a solution, in this case,
$$
z=1+2 \times 7=15 \text {. }
$$

When $x \geqslant 6$, $2^{x-2}-7^{y}=1$, taking both sides modulo 16 gives
$$
7^{y} \equiv 15(\bmod 16) \text {. }
$$

However, $7^{2} \equiv 1(\bmod 16)$, so for any positive integer $y$,
$$
7^{y} \equiv 1,7(\bmod 16),
$$

there is no solution.
In summary, the complete solution to the indeterminate equation is $(5,1,15)$.