Let $S$ denote the unique set of integers such that \[\sum \limits_{n \in S} 2^n = \sqrt{d}\] Let the number of digits of $\sqrt{d}$ before the decimal point be $\chi$. As a result, $\sqrt{d} < 2^{\chi+1}$. Consider the number $v$ obtained by taking the first $\chi+\ell$ digits of $\sqrt{d}$, where $\ell \geq \chi$. Note that \[v = \frac{ \lfloor 2^{\ell} \sqrt{d} \rfloor}{2^{\ell}} = \sqrt{d} - \frac{ \{ 2^{\ell} \sqrt{d} \} }{2^{\ell}}\]We have \[v^2 = d + \frac{\{2^{\ell} \sqrt{d}\}^2}{4^{\ell}} - \frac{\{2^{\ell} \sqrt{d}\} \sqrt{d} }{2^{\ell-1}}\] In binary, $d$ has no digits after the decimal point. The second term above is strictly less than $1/4^{\ell}$, and hence in its binary, the first $2\ell$ digits after the decimal point are zero. The third term is less than $\frac{1}{2^{\ell-\chi-2}}$, and thus in its binary, the first $\ell-\chi-2$ digits are zero. Since $2\ell \geq \ell-\chi-2$ and the second term is smaller than the third, it follows that $v^2$ has at least $\ell-\chi-2$ bits after the decimal point set to one. We will be now using the following lemma: Lemma: Suppose that $m$ is a rational number that has a finite binary expansion. Further, let $S$ be a multiset of integers such that \[\sum \limits_{n \in S} 2^n = m\]Then $|S| \geq s_2(m)$. Proof: This is easy to prove. While $S$ has two equal elements, we may combine them and reduce the size of $S$ by one. Since the final set $S$ must exactly be the binary expansion of $m$, we obtain that $|S| \geq m$, as claimed. Note that $s_2(v^2) \geq \ell-\chi-2$. Moreover, we may write $v^2$ as: \[\sum \limits_{u \in S, u \geq -\ell} 2^{2u} + \sum \limits_{u > v \geq -\ell, u, v \in S} 2^{u+v+1}\]We are using exactly $\binom{s(\chi+\ell) + 1}{2}$ powers of two in the representation above. From the lemma, we therefore obtain \[\binom{s(\chi+\ell) + 1}{2} \geq \ell-\chi-2\]Write $N = \chi + \ell$. The above yields \[(s(N)+1)^2 > 2N-4\chi-2\]This can easily be seen to imply $s(N) > \sqrt{2N} - 2$ for all large $N$. $\square$

In short, you can cut off the representation of $\sqrt d$ after $n$ decimals, and suppose there are $k$ digits which are $1$. Now represent $\sqrt d$ as a sum $s$ of $k$ powers of $2$, plus some residue which is smaller than $2^{c_1-n}$. Squaring, you get that $d$ is equal to $s^2$ $+$ some stuff which is less than $2^{c_2-n}$ for some constants $c_1,c_2$. Since something smaller than $2^{c_2-n}$ completes $s^2$ to an integer, then $s^2$ must have its first $n-c_2$ decimals equal to $1$. This is enough to finish since $s^2$ is the sum of $\frac{k^2+k}2$ powers of 2, so $\frac{k^2+k}2 \ge n-c_2\implies k>\sqrt{2n}-1$.