1. Lemma Statement and Setup:
We need to prove that for a given constant $\lambda$, the set of powers of $2$ with a 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.

2. Proof of Lemma:
Assume to the contrary that the digits $c_{k+1}, c_{k+2}, \cdots, c_{4k}$ 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 $2^n \equiv t \pmod{10^{4k}}$.

An easy induction shows that $n \geq m$ and so $n \geq 4k$. Thus $t \equiv 2^n \equiv 0 \pmod{2^{4k}}$, so $2^{4k}$ divides $t$ and $t \geq 16^k$. But $t < 10^k < 16^k$, which is a contradiction. Therefore, our assumption that all digits $c_{k+1}, c_{k+2}, \cdots, c_{4k}$ are zero is false.

3. Application of Lemma:
Now consider a sufficiently large $N$. Since the last digit of $2^N$ is at least $2$, we can start the process of searching for 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 this non-zero digit 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.

4. Conclusion:
For some $N$ (it is not hard to see that $N \geq 4^{\lambda-1}$ will do the trick), all the powers of $2$ after $2^N$ will have a digit sum greater than $\lambda$. Therefore, the number of powers of $2$ with a digit sum smaller than $2019^{2019}$ is finite.

$\blacksquare$