Example 9 Find the positive integer solutions $(x, y, z, t)$ of the indeterminate equation $1+5^{x}=2^{y}+2^{x} \cdot 5^{t}$.
Let $(x, y, z, t)$ be a positive integer solution to the original equation. Taking both sides of the equation modulo 5, we have $2^{y} \equiv 1(\bmod 5)$. Since 2 is a primitive root modulo 5, it follows that $4 \mid y$. At this point, taking both sides of the equation modulo 4, we get $2^{z} \equiv 2(\bmod 4)$, so $z=1$.

We set $y=4r$, yielding
$$5^{x}-2 \cdot 5^{t}=16^{r}-1$$

Taking both sides of (12) modulo 3, we have $(-1)^{x}-(-1)^{t+1} \equiv 0(\bmod 3)$, hence
$$x \equiv t+1(\bmod 2)$$

Further, taking both sides modulo 8, we get $5^{x} \equiv 2 \cdot 5^{t}-1 \equiv 1(\bmod 8)$ (using $5 \equiv 5$ or $1(\bmod 8)$). Thus, $x$ is even, and $t$ is odd.

Notice that if $t=1$, then $5^{x}=16^{r}+9$. Setting $x=2m$, we have
$$\left(5^{m}-3\right)\left(5^{m}+3\right)=16^{r}$$

Since $\left(5^{m}-3,5^{m}+3\right)=\left(5^{m}-3,6\right)=2$, it follows that
$$5^{m}-3=2, \quad 5^{m}+3=2^{4r-1},$$

thus $m=1, r=1$. If $t>1$, then $5^{3} \mid\left(5^{x}-2 \cdot 5^{t}\right)$, and hence $5^{3} \mid\left(16^{r}-1\right)$. By the binomial theorem, expanding $16^{r}-1=(15+1)^{r}-1$, we see that $5 \mid r$. Thus, we can set $r=5k$, yielding
$$16^{r}-1=16^{5k}-1 \equiv\left(5^{5}\right)^{k}-1 \equiv\left(5 \times 3^{2}\right)^{k}-1 \equiv 0 \quad(\bmod 11)$$

This requires $11 \mid\left(5^{x}-2 \cdot 5^{t}\right)$, hence $11 \mid\left(5^{x-t}-2\right)$. However, for any $m \in \mathbf{N}^{*}$, we have $5^{n}=5,3,4,9$ or $1(\bmod 11)$, and it is impossible for
$$5^{x-t} \equiv 2(\bmod 11)$$

to occur, leading to a contradiction.
In summary, the equation has a unique solution
$$(x, y, z, t)=(2,4,1,1)$$