There's an elementary way of solving this (and similar equations). Let's start with $$ 1+2^n=5^a(1+10^m) $$ which you want to solve in positive integers $n$, $a$, $m$. Clearly $m < n$ and $a < n$. As wccanard points out, $2$ is a primitive root modulo $5^a$ and so we obtain that $n$ is divisible by $2 \cdot 5^{a-1}$. In particular, $$ 2 \cdot 5^{a-1} \le n. $$ Now let's use the fact that $m < n$ are reduce the equation modulo $2^m$. We obtain, $$ 5^a \equiv 1 \pmod{2^m}. $$ As in your question you write this as $$ 4a + 16 \binom{a}{2}+\cdots \equiv 0 \pmod{2^m}. $$ This implies that $2^{m-2} \mid a$, and so $$ 2^{m-2} \le a. $$ The inequalities we now have show that the left-hand side of the equation is much bigger than the right-hand side as soon as the $n$ is large!

Appended by the OP: With apologies to Samir if he had something slicker in mind, I thought I'd add my own elaboration on when $5^a(1+10^m)$ becomes bigger than $(1+2^n)$.

The first displayed inequality implies $5^a \le (5/2)n$ and the second, combined with $a \lt n$, gives $2^m \le 4a < 4n.$ Clearly, $1+10^m \lt 16^m = (2^m)^4\lt (4n)^4$. This all gives

$$2^n \lt 1+2^n = 5^a(1+10^m) \lt (5/2)n(4n)^4 = 640n^5,$$

and it's easy to check that this implies $n\lt 35$.

Knowing also that $n$ must be congruent to 2 mod 4 means we can finish the problem off by checking the factorizations of $1+2^n$ for $n=2,6,10,14,18,22,26,30,$ and $34$, which is easy enough to do. A less crude estimate than $1+10^m \lt 16^m$ might shave off a few of the larger values of $n$, but it doesn't seem worth the effort.

There's an elementary way of solving this (and similar equations). Let's start with $$ 1+2^n=5^a(1+10^m) $$ which you want to solve in positive integers $n$, $a$, $m$. Clearly $m < n$ and $a < n$. As wccanard points out, $2$ is a primitive root modulo $5^a$ and so we obtain that $n$ is divisible by $2 \cdot 5^{a-1}$. In particular, $$ 2 \cdot 5^{a-1} \le n. $$ Now let's use the fact that $m < n$ are reduce the equation modulo $2^m$. We obtain, $$ 5^a \equiv 1 \pmod{2^m}. $$ As in your question you write this as $$ 4a + 16 \binom{a}{2}+\cdots \equiv 0 \pmod{2^m}. $$ This implies that $2^{m-2} \mid a$, and so $$ 2^{m-2} \le a. $$ The inequalities we now have show that the left-hand side of the equation is much bigger than the right-hand side as soon as the $n$ is large!

Appended by the OP: With apologies to Samir if he had something slicker in mind, I thought I'd add my own elaboration on when $(1+2^n)$ becomes bigger than $5^a(1+10^m)$.

The first displayed inequality implies $5^a \le (5/2)n$ and the second, combined with $a \lt n$, gives $2^m \le 4a < 4n.$ Clearly, $1+10^m \lt 16^m = (2^m)^4\lt (4n)^4$. This all gives

$$2^n \lt 1+2^n = 5^a(1+10^m) \lt (5/2)n(4n)^4 = 640n^5,$$

and it's easy to check that this implies $n\lt 35$.

Knowing also that $n$ must be congruent to 2 mod 4 means we can finish the problem off by checking the factorizations of $1+2^n$ for $n=2,6,10,14,18,22,26,30,$ and $34$, which is easy enough to do. A less crude estimate than $1+10^m \lt 16^m$ might shave off a few of the larger values of $n$, but it doesn't seem worth the effort.

There's an elementary way of solving this (and similar equations). Let's start with $$ 1+2^n=5^a(1+10^m) $$ which you want to solve in positive integers $n$, $a$, $m$. Clearly $m < n$ and $a < n$. As wccanard points out, $2$ is a primitive root modulo $5^a$ and so we obtain that $n$ is divisible by $2 \cdot 5^{a-1}$. In particular, $$ 2 \cdot 5^{a-1} \le n. $$ Now let's use the fact that $m < n$ are reduce the equation modulo $2^m$. We obtain, $$ 5^a \equiv 1 \pmod{2^m}. $$ As in your question you write this as $$ 4a + 16 \binom{a}{2}+\cdots \equiv 0 \pmod{2^m}. $$ This implies that $2^{m-2} \mid a$, and so $$ 2^{m-2} \le a. $$ The inequalities we now have show that the left-hand side of the equation is much bigger than the right-hand side as soon as the $n$ is large!

There's an elementary way of solving this (and similar equations). Let's start with $$ 1+2^n=5^a(1+10^m) $$ which you want to solve in positive integers $n$, $a$, $m$. Clearly $m < n$ and $a < n$. As wccanard points out, $2$ is a primitive root modulo $5^a$ and so we obtain that $n$ is divisible by $2 \cdot 5^{a-1}$. In particular, $$ 2 \cdot 5^{a-1} \le n. $$ Now let's use the fact that $m < n$ are reduce the equation modulo $2^m$. We obtain, $$ 5^a \equiv 1 \pmod{2^m}. $$ As in your question you write this as $$ 4a + 16 \binom{a}{2}+\cdots \equiv 0 \pmod{2^m}. $$ This implies that $2^{m-2} \mid a$, and so $$ 2^{m-2} \le a. $$ The inequalities we now have show that the left-hand side of the equation is much bigger than the right-hand side as soon as the $n$ is large!

Appended by the OP: With apologies to Samir if he had something slicker in mind, I thought I'd add my own elaboration on when $5^a(1+10^m)$ becomes bigger than $(1+2^n)$.

The first displayed inequality implies $5^a \le (5/2)n$ and the second, combined with $a \lt n$, gives $2^m \le 4a < 4n.$ Clearly, $1+10^m \lt 16^m = (2^m)^4\lt (4n)^4$. This all gives

$$2^n \lt 1+2^n = 5^a(1+10^m) \lt (5/2)n(4n)^4 = 640n^5,$$

and it's easy to check that this implies $n\lt 35$.

Knowing also that $n$ must be congruent to 2 mod 4 means we can finish the problem off by checking the factorizations of $1+2^n$ for $n=2,6,10,14,18,22,26,30,$ and $34$, which is easy enough to do. A less crude estimate than $1+10^m \lt 16^m$ might shave off a few of the larger values of $n$, but it doesn't seem worth the effort.