Is there an expression which defines $y^2$ with $y \in \mathbb{N}$ into as little as possible sums of $2^x$ where $x \in \mathbb{N}_0$?
i.e.
$1^2 = 2^0$
$2^2 = 4 = 2^2$
$3^2 = 9 = 2^3 + 2^0$
$4^2 = 16 = 2^4$
$5^2 = 25 = 2^4 + 2^3 + 2^0$
$6^2 = 36 = 2^5 + 2^2$
$7^2 = 49 = 2^5 + 2^4 + 2^0$
$8^2 = 64 = 2^6$
$9^2 = 81 = 2^6 + 2^4 + 2^0$
$10^2 = 100 = 2^6 + 2^5 + 2^2$
$11^2 = 121 = 2^6 + 2^5 + 2^4 + 2^3 + 2^0$
$12^2 = 144 = 2^7 + 2^4$
$13^2 = 169 = 2^7 + 2^5 + 2^3 + 2^0$
$14^2 = 196 = 2^7 + 2^6 + 2^2$
$15^2 = 225 = 2^7 + 2^6 + 2^5 + 2^0$
$16^2 = 256 = 2^8$
$17^2 = 289 = 2^8 + 2^5 + 2^0$
$18^2 = 324 = 2^8 + 2^6 + 2^2$
$19^2 = 361 = 2^8 + 2^6 + 2^5 + 2^3 + 2^0$
$20^2 = 400 = 2^8 + 2^7 + 2^4$
To any natural $m$ you associate the sequence
$$p_n:=\lfloor m\,2^{-n}\rfloor\bmod2,$$ where the term $p_n$ indicates if the power $2^n$ belongs to the sum (as of $2^n>m, p_n=0$).
E.g. $25\to1,0,0,1,1,0,0,0,0,0,\cdots\equiv 2^0+2^3+2^4$.
I don't know of a direct formula that would give the number of terms, nor the powers of these terms.
Update:
The number of terms is known as the Hamming weight of the number.