Three, (50 points) Find all positive integer solutions $(x, y, z, t)$ of the equation $2^{y}+2^{z} \times 5^{t}-5^{x}=1$.

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Three, Solution: If $(x, y, z, t)$ is a positive integer solution to the equation, then $2^{y} \equiv 1 \pmod{5}$.
Since $2^{4} \equiv 1 \pmod{5}$, it follows that $4 \mid y$.
Taking the equation modulo 4, we get $2^{x} \equiv 2 \pmod{4}$, hence $z=1$.
Let $y=4r$, then $5^{x}+1=2^{4r}+2 \times 5^{t}$, which simplifies to $5^{x}-2 \times 5^{t} = 16^{r}-1$. Taking both sides modulo 3, we get $(-1)^{x}+(-1)^{t} \equiv 0 \pmod{3}$, so $x, t$ are one odd and one even.

Since $5^{t} \equiv 1,5 \pmod{8}$, taking both sides of $5^{x}=2 \times 5^{t}+16^{r}-1$ modulo 8, we get
$5^{x} \equiv 2 \times 5^{t}-1 \equiv 1 \pmod{8}$, hence $x$ is even and $t$ is odd.
(1) If $t=1$, then $5^{x}=16^{r}+9$. Let $x=2m$, we have $(5^{m}-3)(5^{m}+3)=16^{r}$,
$$
\begin{array}{l}
\quad \text { By } (5^{m}-3,5^{m}+3)=(5^{m}-3,6)=2 \Rightarrow 5^{m}-3= \\
2,5^{m}+3=2^{4r-1}, \\
\quad \Rightarrow m=1,2^{4r-1}=2^{3} \Rightarrow r=1, y=4, x=2 \Rightarrow (x, y, \\
z, t)=(2,4,1,1) .
\end{array}
$$
(2) If $t>1$, then $t \geq 3, x \geq 4$,
$5^{3}(5^{x}-2 \times 5^{t})=16^{r}-1$. Since $16^{r}-1=(15+1)^{r}-1=15^{r}+\cdots+\binom{r}{2}15^{2}+\binom{r}{1}15$, it follows that $5 \mid r$.

Let $r=5k$, then $16^{5k}-1 \equiv 5^{5k}-1 \equiv (5 \times 3^{2})^{k}-1 \equiv 0 \pmod{11}$,

Thus, $11 \mid (5^{x}-2 \times 5^{t})=5^{t}(5^{x-t}-2) \Rightarrow 11 \mid (5^{x-t}-2)$,

But $5^{n} \equiv 1,3,4,5,9 \pmod{11}$, meaning 11 cannot divide $5^{x-t}-2$, a contradiction.
Therefore, the original equation has a unique solution $(x, y, z, t)=(2,4,1,1)$.