I suppose this will follow from the plausible Cramer's conjecture about prime gap of $O(\log^2{p_n})$.
Let $n=(2^k-1)2^m$ with $m < k$ and $2^m>C\log^2{n}$ (actually $m > \log{(C\log^2({2^m(2^k-1))}}/\log{(2)}$ will do.)
$n$ has $k$ ones and the (much) less zeros. The interval $(n,n+C\log^2{n})$ with $2^m>C\log^2{n}$ will contain a prime $p=n+\delta$. $\delta$ will contribute at least one $1$ to the zero bits of $n$ keeping all of the ones. There are infinitely many choices for $m,k$.
Not m,k$producing distinct primes.. Legendre conjecture probably will do too. Note that the doubly logarithmic choice of$m$contributes relatively few zeros. Legendre Added So if you believe Cramer's conjecture probably will do too , there are infinitely many$n$for which$n$bit primes have only$O(\log{n})$zeros in their Binary expansion. (btw, I would be very interested in unconditional answer to the question). 2 Added 2^m> I suppose this will follow from the plausible Cramer's conjecture about prime gap of$O(\log^2{p_n})$. Let$n=(2^k-1)2^m$with$m < k$and$2^m>C\log^2{n}$(actually$ m > \log{(C\log^2({2^m(2^k-1))}}/\log{(2)} $will do.)$n$has$k$ones and the interval$(n,n+C\log^2{n})$with$2^m>C\log^2{n}$will contain a prime$p=n+\delta$.$\delta$will contribute at least one$1$to the zero bits of$n$keeping all of the ones. There are infinitely many choices for$m,k$. Not that the doubly logarithmic choice of$m$contributes relatively few zeros. Legendre conjecture probably will do too . 1 I suppose this will follow from the plausible Cramer's conjecture about prime gap of$O(\log^2{p_n})$. Let$n=(2^k-1)2^m$with$m < k$(actually$ m > \log{(C\log^2({2^m(2^k-1))}}/\log{(2)} $will do.)$n$has$k$ones and the interval$(n,n+C\log^2{n})$with$2^m>C\log^2{n}$will contain a prime$p=n+\delta$.$\delta$will contribute at least one$1$to the zero bits of$n$keeping all of the ones. There are infinitely many choices for$m,k$. Not that the doubly logarithmic choice of$m\$ contributes relatively few zeros.