Example 15 Find the positive integer solutions $(x, y, z, t)$ of the indeterminate equation $1+5^{x}=2^{y}+2^{z} \cdot 5^{t}$.
Let $(x, y, z, t)$ be a positive integer solution of the original equation.
Taking 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 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 the above equation modulo 3, we have $(-1)^{x}-(-1)^{t+1} \equiv 0(\bmod 3)$, hence $x \equiv t+1(\bmod 2)$. Further, taking the equation modulo 8, we get $5^{x} \equiv 2 \cdot 5^{t}-1 \equiv 1(\bmod 8)$ [here we use $5^{t} \equiv 5$ or $1(\bmod 8)$], so $x$ 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 $5^{m}-3=2$ and $5^{m}+3=2^{4r-1}$, thus $m=1$ and $r=1$; if $t>1$, then $5^{3} \mid 5^{x}-2 \cdot 5^{t}$, hence $5^{3} \mid 16^{r}-1$. By the binomial theorem, expanding $16^{r}-1=(15+1)^{r}-1$, we see that $5 \mid r$, so we set $r=5k$, yielding $16^{r}-1=16^{5k}-1 \equiv (5^{5})^{k}-1 \equiv (5 \times 3^{2})^{k}-1 \equiv 0(\bmod 11)$. This requires $11 \mid 5^{x}-2 \cdot 5^{t}$, hence $11 \mid 5^{x-t}-2$. However, for any $m \in \mathbf{N}^{*}$, we have $5^{n} \equiv 5, 3, 4, 9$ or $1(\bmod 11)$, and it is impossible for $5^{x-t} \equiv 2(\bmod 11)$, leading to a contradiction.
In conclusion, the equation has a unique solution $(x, y, z, t)=(2,4,1,1)$.