Solution:
Call a number good if all of its positive divisors have at most two $1$'s in their binary representations. Then, if $p$ is an odd prime divisor of a good number, $p$ must be of the form $2^{k}+1$. The only such primes less than $2012$ are $3, 5, 17$, and $257$, so the only possible prime divisors of $n$ are $2, 3, 5, 17$, and $257$.

Next, note that since $(2^{i}+1)(2^{j}+1) = 2^{i+j} + 2^{i} + 2^{j} + 1$, if either $i$ or $j$ is greater than $1$, then there will be at least $3$ $1$'s in the binary representation of $(2^{i}+1)(2^{j}+1)$, so $(2^{i}+1)(2^{j}+1)$ cannot divide a good number. On the other hand, if $i = j = 1$, then $(2^{1}+1)(2^{1}+1) = 9 = 2^{3}+1$, so $9$ is a good number and can divide a good number. Finally, note that since multiplication by $2$ in binary just appends additional $0$'s, if $n$ is a good number, then $2n$ is also a good number.

It therefore follows that any good number less than $2012$ must be of the form $c \cdot 2^{k}$, where $c$ belongs to $\{1, 3, 5, 9, 17, 257\}$ (and moreover, all such numbers are good). It is then straightforward to check that the largest such number is $1536 = 3 \cdot 2^{9}$.