The answer is no. Assume such a $k$ exists and let $N$ be a large integer. Then every number from $\{1,2,\dots,N\}$ can be written as $a_1^2+a_2^4+\dots+a_k^{2^k}$. Each such representation must have $a_1^2\leq N$, i.e. $0\leq a_1\leq\sqrt N$. Hence, there are at most $\sqrt N+1$ choices for $a_1$. Analogously, there are at most $\sqrt[4]N+1$ choices for $a_2$, etc. All in all, this shows that we have at most $\prod\limits_{j=1}^k\left(N^{2^{-j}}+1\right)$ possibilities. Since each number in $\{1,2,\dots,N\}$ must have such a representation, this must be at least $N$. However, multiplying out all terms shows that the largest power of $N$ which occurs is $N^{\frac 12+\frac 14+\dots+\frac{1}{2^k}}=N^{1-\frac{1}{2^k}}$. Hence, this function grows slower than the function $N\mapsto N$, resulting in a contradiction if we take $N$ sufficiently large.

The same result (and proof idea) holds more generally when the exponents $b_1, b_2, \ldots, b_k$ are such that $\sum_{i=1}^k \frac{1}{b_i} < 1$. (In our case $\sum_{i=1}^{k} \frac{1}{2^i} = 1 - \frac{1}{2^k}$).