Sketch: Infinite. Let $S_n$ be the set positive integers less than or equal to $n$ that are $3 \pmod{4}$ and can be written as the sum of two powers. The $3 \pmod{4}$ condition prohibits sums of two squares. We can estimate the number of $k-th$ powers less than or equal to $n$ as $n^{\frac{1}{k}}$ and therefore we can estimate the size of $S$ to be at most $$\sum_{i, j, (i, j) \neq (2, 2)} n^{\frac{1}{i} + \frac{1}{j}}$$ The largest exponent in this sum is $\frac{5}{3}$, so we expect this to be smaller than, say, $0.24n$ for sufficiently large $n$, which is enough to imply the result. The actual solution requires a bit more care as we need to prevent counting the pair $1^k + 1^{\ell} = 2$ lots of times, but its not too difficult to do the bounding.

I guess it's simpler: Suppose that exist $N$ such that all numbers greater than $N$ can be written as sum of two powers. First, observe that the number of powers less than or equal to $n$ may be counted as the following: $n^{1/2} + n^{1/3} + ... + n^{1/log_2 n} < log_2 n \times n^{1/2}$ In the same way, the number of odd powers can be upper bounded as: $n^{1/3} + n^{1/5} + .... + n^{1/log_2 n} < log_2 n \times n^{1/3}$ This implies that the number of pairs of powers such that at least one of them is an odd exponent is at most: $log_2 n \times n^{1/2} \times log_2 n \times n^{1/3} = (log_2 n)^2 \times n^{5/6}$. However, all numbers of the form $4k+3$ from a certain point are represented as sum of 2 powers with at least one odd among the 2 exponents. Then we would have $(log_2 n)^2 \times n^{5/6}$ being bigger than $\dfrac{n}{4} - K$ where K is a constant equals the number of guys less than $N$ that are $3(mod4)$ which can't be represented as sum of 2 powers. This is clearly an absurd.