Just to be concrete, consider the digits to be binary. Hasse showed that among all the primes, only a fraction of $17/24 < 1$ divide a number of the form $2^n+1$. As a result, the integers that divide a number with just two non-zero digits have zero density.

On the other hand, since $A + A = \mathbb{F}_p$ for a typical set $A \subset \mathbb{F}_p$ of cardinality, say, $> p/\log{p}$ (and much lower than that), and because under GRH and with probability $1$ there are at least as many powers of $2$ mod $p$, we *expect* a full density of the primes to divide a number of the form $2^m+2^n+1$. [Added: As shown by Skalba in the reference provided by "so-called friend Don," we can actually prove that this is true for all primes $p \gg_{\epsilon} 0$ having $r := \mathrm{ord}_p^{\times}2 > p^{\frac{3}{4}+\epsilon}$: consider Weil's bound for the Fermat curve of degree $(p-1)/r$. This applies also to the linked question. ]

So it would seem legitimate to ask if the probability might be positive for a random integer to have a multple composed of only three non-zero digits. Note that $2^k-1$ does not have this property for $k > 3$ (all its multiples have digit sum exceding $ck$), and this family already furnishes an infinite set of pairwise co-prime integers without the property.

Yet I thought it would be hard to believe that a random integer, with positive probability, will have a multiple with bounded digit sum. Is there a (better) heuristic for the expected number up to a given bound $X$ of the positive integers having a multiple of the form $2^m+2^n+1$? On the opposite extreme, wouldn't most $N$ require, for all their multiples, as many as $(1+o(1))\frac{\log{N}}{2\log{2}}$ binary ones?