I saw the solution:
Suppose $s \le r$. If $s < r$ then $2^s \le 2^r$. Since the number of digits in $2^s$and $2^r$ are the same, we have $2^r < 10 * 2^s < 2^{s+4} $. Thus we have $2^s < 2^r <2^{s+4}$ which gives $r = s+1, s+2, s+3 $ . Since $2^r$ is obtained from $2^s$ by permuting its digits, $2^r - 2^s$ is divisible by $9$. If $r = s+1, s+2, s+3$, clearly, $2^r$ is not divisible by 9. So, we conclude that $s < r$ is not possible. Therefore, $s = r$.
Everything is good. But I cannot understand the divisible by 9 thing which I have highlighted too.
If you have a natural number $m$ and if $s$ is the sum of its digits, then $m\equiv s\pmod 9$. Therefore, if $n$ is another natural number obtained permuting the digts of $m$, then $n\equiv s\pmod9$ too, and therefore $m-n\equiv0\pmod 9$. In other words, $9\mid m-n$.