, 2011

Suppose that the digits of $2^a$ can be rearranged to form $2^b$, with $a > b$. Then, since the two numbers have the same digit set, it follows that they're congruent modulo $9$. Hence $9 \mid 2^a - 2^b = 2^b(2^{a-b} - 1)$ and so $2^{a-b} \equiv_9 1$. However, the smallest positive power of $2$ with this property is $2^6$, so $a-b \ge 6$. But in that case, $2^a \ge 2^{b+6} = 64 \cdot 2^b > 10 \cdot 2^b$, which means that $2^a$ and $2^b$ can't have the same number of digits, a contradiction. So no such pair exists.