Integer factorization still hard with Hamming weight hypothesis? Consider the following problem:

Factorize a $n$-bit integer $c$ knowing that it is the product of two integers with known Hamming weight $h$.

Is there a way to prove that this is still hard? I have parameters $n=1024$ and $h=80$, but am willing to augment $h$ if this turns out to be easy. I think this problem is current-factorization-algorithms-related, since I did not find an easy way to describe the distribution of these type of numbers (my try considered quadratic expressions of dependent Bernoulli trials). So the question may be reformulated: Does this hypothesis helps any of the actual factorization algorithms?
I suspect that this might be vulnerable to a combinatorial factoring attack.

In this attack, we look at possible solutions to $pq = n \bmod 2^k$, and then extend $p$ and $q$ one bit to list the possible solutions to $pq = n \bmod 2^{k+1}$

Now, if we have no further information about $p$ and $q$, this turns out to be no more efficient than brute force search (as at each step, the number of possibilities doubles, and we have no way to distinguish likely versus unlikely possibilities).

However, in your scenario, we know that both $p$ and $q$ are strongly biased toward 0 bits, and so it would make sense to order our search to explore the possibilities with $p$ and $q$ being mostly 0 first.  The immediately obvious approach would be to keep a list of possible solutions to $pq = n \bmod 2^k$, ordered by hamming weight of $p$ and $q$ (and $k$ somehow); and explore the long solutions with low hamming weight first.

It would seem plausible that this might quickly find a solution if the hamming weight was sufficiently small.  How well would it work in practice (and how small is 'small')?  I suspect that's something that would need to be experimented with.