Brazilian Mathematical Olympiad

Take $n$ to be the smallest integer such that $2^n \ge 10^{m-1}$. Then $2^{n-1} < 10^{m-1}$, so $2^{n+2} < 8 \cdot 10^{m-1} < 10^m$. So $2^n$, $2^{n+1}$ and $2^{n+2}$ all have $m$ digits. Thus there are at least $3$ powers of $2$ with $m$ digits.

$2^{n-1} \ge 5 \cdot 10^{m-2}$ (otherwise $2^n < 10^{m-1}$). Hence $2^{n+4} = 32 \cdot 5 \cdot 10^{m-2} > 10^m$, so $2^{n+4}$ has more than $m$ digits. Thus there are at most $4$ powers of $2$ with $m$ digits.

There are $4$ if there is an integer between $\frac{m-1}{\log_{10} 2}$ and $\frac{m}{\log_{10} 2} - 3$.