This problem is equivalent to finding all integer solutions to the following equation where $a$ is a single digit not equaling to $0$: $$2^n=2^m+a\times 10^k$$

This equation can be re-written as $$2^m(2^{n-m}-1)=a\times 10^k$$

Because the original number should contain at least two digits, it is clear that $k$ must be greater or equal to $1$.

If $k > 1$, then $5^2\mid\left(a\times 10^k\right)$. This means that the right side must be a multiple of $5^2$. Because $2^m\not\mid 5^2$, it must hold that $$5^2\mid (2^{n-m}-1)\implies 2^{n-m}\equiv 1\pmod{5^2}$$

As the multiplicative order of $2$ modulo $5^2$ is $20$, it must hold that $20\mid (n-m)$.

It follows that $(2^{20}-1)$ divides $(2^{n-m}-1)$. Furthermore, because $(2^5 -1) \mid (2^{20}-1)$, it must be true that $(2^{n-m}-1)$ is a multiple of $(2^5-1)=31$. This means that the left side of the second equation above is a multiple of $31$. But it is clear that the right side of that equation cannot be a multiple of $31$. Hence, we conclude $k > 1$ cannot hold. When $k=1$, the original number must be a two-digit number. Examining all possible candidates find only two solutions: $$\boxed{32}\quad\text{and}\quad\boxed{64}$$