Use the fact that $9$ divides $2^r-2^s$. If $r>s$ this means that $9$ divides $2^{r-s}-1$. The smallest difference $r-s$ which satisfies this is $r-s=6$. Thus $2^r \geq 64 2^s$, so $2^r,2^s$ do not have the same number of digits.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=616666

Oh I thought you missed this problem. Sorry! I am deleting this thread.

Without loss of generality assume that $r>s$. Let $r = s+k$, where $k>0$. As $2^{r}$ and $2^s$ have same number of digits $k<4$(since $k>4$ will imply multiplying $2^s$ by a number greater than $10$ and thus ending up with a greater number of digits). Now, as they have same number of digits their sum of digits is same so, they are congruent modulo $9$. So $2^{s+k} - 2^{s} = 2^{s}(2^k - 1) \equiv 0$ modulo $9$. So since $(9,2^s) = 1$(they are co-prime) this implies that $2^k - 1 \equiv 0$ modulo $9$. $k<4$ implies that $2^k - 1$ can take values $1,3,7$ or all of which leads to a contradiction. So $k=0$ which gives $r=s$

Its dissapointing to observe that this olympiad problem was already present in the popular book 'Solving Mathematical problems a personal approach' Terence Tao ( problem 2.2 )

Submathematics wrote: Its dissapointing to observe that this olympiad problem was already present in the popular book 'Solving Mathematical problems a personal approach' Terence Tao ( problem 2.2 ) Its also present in $mathematical$ $olympiad$ $treasures$ by Titu Andreescu