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

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
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: $2^{z} \equiv 2(\bmod 4)$, hence $z=1$.
Let $y=4 r$, then $2^{4 r}+2 \times 5^{t}=1+5^{x}$, which simplifies to $5^{x}-2 \times 5^{t}=16^{r}-1$.
Taking the above equation modulo 3, we get: $(-1)^{x}+(-1)^{t} \equiv 0(\bmod 3)$, so $x, t$ are one odd and one even.
Since $5^{t} \equiv 1$ or $5(\bmod 8)$, taking $5^{x}=2 \times 5^{t}+16^{r}-1$ modulo 8, we get
$5^{x} \equiv 2 \times 5^{t}-1 \equiv 1(\bmod 8)$, hence $x$ is even and $t$ is odd.
(1) If $t=1$, then $5^{x}=16^{r}+9$, let $x=2 m$, we have
$\left(5^{m}-3\right)\left(5^{m}+3\right)=16^{r}$, and the greatest common divisor of $5^{m}-3$ and $5^{m}+3$ is
$\left(5^{m}-3,5^{m}+3\right)=\left(5^{m}-3,6\right)=2$,
thus $5^{m}-3=2,5^{m}+3=2^{4-1}$, so $m=1, r=1,(x, y, z, t)=(2,4,1,1)$.
(2) If $t>1$, then $t \geqslant 3, x \geqslant 4$, then from $5^{x}-2 \times 5^{t}=16^{r}-1$ we get: $5^{3} \mid\left(5^{x}-2 \times 5^{t}\right)$,
which means $5^{3} \mid\left(16^{r}-1\right)$, and since $16^{r}-1=(15+1)^{r}-1=15^{r}+\cdots+C_{r}^{2} 15^{2}+C_{r}^{1} 15$,

we have $5 \mid r$, let $r=5 k$, then $16^{5 k}-1 \equiv 5^{5 k}-1 \equiv\left(5 \times 3^{2}\right)^{k}-1 \equiv 0(\bmod 11)$,
which implies $11 \mid\left(5^{x}-2 \times 5^{t}\right)=5^{t}\left(5^{x-t}-2\right)$, hence $11 \mid\left(5^{x-t}-2\right)$,
but $5^{n} \equiv 1,3,4,5,9(\bmod 11)$, so 11 does not divide $\left(5^{x-t}-2\right)$, a contradiction.
Therefore, the original equation has a unique solution $(x, y, z, t)=(2,4,1,1)$.