Solution:
Answer: 5 Clearly we may biject squarely sets with binary representations of perfect squares between 1 and $2^{0}+\cdots+2^{8}=2^{9}-1=511$, so there are 22 squarely sets, corresponding to $n^{2}$ for $n=1,2, \ldots, 22$. For convenience, we say $N$ is (super) squarely if and only if the set corresponding to $N$ is (super) squarely.
The general strategy is to rule out lots of squares at a time, by searching for squares with few missing digits (and ideally most 1's consecutive, for simplicity). We can restrict ourselves (for now) to odds; $(2k)^{2}$ is just $k^{2}$ with two additional zeros at the end. $1,9,25,49,81$ are ineffective, but $121=2^{7}-7=2^{6}+2^{5}+2^{4}+2^{3}+2^{0}$ immediately rules out all odd squares up to $9^{2}$, as they must be $1\pmod{8}$.
Fortunately, $22^{2}=4 \cdot 11^{2}$ is in our range (i.e. less than 512), ruling out all even squares up to $20^{2}$ as well.
This leaves us with $11^{2}, 13^{2}, 15^{2}, 17^{2}, 19^{2}, 21^{2}, 22^{2}$, with binary representations 001111001, 010101001, $011100001, 100100001, 101101001$ (kills $17^{2}$), 110111001 (kills $13^{2}$), 111100100 (kills nothing by parity). Thus $11^{2}, 15^{2}, 19^{2}, 21^{2}, 22^{2}$ are the only super squarely numbers, for a total of 5.