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!