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

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Let $(x, y, z, t)$ be a positive integer solution of the equation. Then $2^{y} \equiv 1(\bmod 5)$.
Since $2^{4} \equiv 1(\bmod 5)$, we have $4 \mid y$.

Taking the equation modulo 4, we get
$$
2^{x}=2(\bmod 4).
$$

Thus, $z=1$.
Let $y=4 r$, then $5^{x}+1=2^{4 r}+2 \times 5^{t}$, which simplifies to
$$
5^{x}-2 \times 5^{t}=16^{r}-1.
$$

Taking the above equation modulo 3, we get
$$
(-1)^{x}+(-1)^{t} \equiv 0(\bmod 3).
$$

Therefore, $x$ and $t$ are of opposite parity.
Since $5^{t}=1$ or $5(\bmod 8)$, taking the equation $5^{x}=2 \times 5^{t}+16^{t}-1$ modulo 8, we get
$$
5^{x} \equiv 2 \times 5^{t}-1 \equiv 1(\bmod 8).
$$

Thus, $x$ is even and $t$ is odd.
(1) If $t=1$, then $5^{x}=16^{r}+9$. Let $x=2 m$, we have
$$
\begin{array}{l}
\left(5^{m}-3\right)\left(5^{m}+3\right)=16^{r}. \\
\text { Since }\left(5^{m}-3,5^{m}+3\right)=\left(5^{m}-3,6\right)=2 \\
\Rightarrow 5^{m}-3=2,5^{m}+3=2^{4 r-1} \\
\Rightarrow m=1,2^{4 r-1}=2^{3} \Rightarrow r=1, y=4, x=2 \\
\Rightarrow(x, y, z, t)=(2,4,1,1).
\end{array}
$$
(2) If $t>1$, then $t \geqslant 3, x \geqslant 4$,
$$
5^{3} \mid\left(5^{x}-2 \times 5^{t}\right)=16^{r}-1.
$$

Since $16^{r}-1=(15+1)^{r}-1=15^{r}+\cdots+\mathrm{C}_{r}^{2} 15^{2}+$ $\mathrm{C}_{r}^{\mathrm{1}} 15$, we have $51 r$.
$$
\begin{array}{l}
\text { Let } r=5 k, \text { then } \\
16^{5 k}-1 \equiv 5^{5 k}-1 \equiv\left(5 \times 3^{2}\right)^{k}-1 \equiv 0(\bmod 11) \\
\Rightarrow 11 \mid\left(5^{x}-2 \times 5^{t}\right)=5^{t}\left(5^{x-t}-2\right) \\
\Rightarrow 11 \mid\left(5^{x-t}-2\right).
\end{array}
$$

However, $5^{n} \equiv 1,3,4,5,9(\bmod 11)$, so $11 \nmid\left(5^{x-4}-2\right)$, which is a contradiction.
Therefore, the original equation has a unique solution $(x, y, z, t)=(2,4,1,1)$.
(Department of Mathematics, Qingdao No. 2 High School, Qingdao, Shandong, 266101)