For a given constant $\lambda, $ we will prove that the set of the powers of $2$ with sum of digits not greater than $\lambda$ is finite. $Lemma:$ Write $2^n=\overline{c_{m}c_{m-1} \cdots c_2c_1}, $ where the $c_i'$s are digits. Consider a digit $c_k,$ which is not equal to zero and $4k \leq m$. Then at least one of ${c_{k+1}, c_{k+2}, \cdots, c_{4k} }$ is not equal to zero. $Proof:$ Assume to the contrary that the digits mentioned above are all equal to zero. Write $t=\overline{c_kc_{k-1} \cdots c_1}$. Now we see that $t<10^k$. We also have $2^n \equiv t(\mod 10^{4k})$. An easy induction shows that $n \geq m$ and so $n \geq 4k$. Thus $t\equiv2^n\equiv0(\mod2^{4k}),$ so $2^{4k}$ divides $t$ and $t\geq16^k$. But $t<10^{k}<16^k, $ contradiction. Now consider a sufficiently large $N$. Since the last digit of $2^N$ is at least $2,$ we can start the process of searching non-zero digits in the expansion of $2^N,$ by choosing $k=1$ in the lemma. This gives us a non-zero digit among $c_2, c_3$ and $c_4$. We take it and use it again in the lemma. This gives another non-zero digit before $c_{17}$. It is evident that by repeating this process we can find as many non-zero digits as we need, so for some $N$ (it is not hard to see that $N\geq4^{\lambda-1}$ will do the trick)$,$ all the powers of $2$ after $2^N$ will have a digit sum greater than $\lambda, \blacksquare$.