I solved it with 2 greedy algorithms: pair up the numbers from lowest to highest power of 2 and from highest to lowest power of 2. Taking the best answer of these 2 algorithms passes the tests, not sure if it is correct. I have a feeling this isn't correct but can't find a counterexample (I also didn't think too hard about it).

Here is a solution. The key observation is that if x = max_ia[i], then x can be only be paired with one other number: 2^k - x, where k is minimal so that 2^k > x. This makes a greedy algorithm clear: sort a[0] < ... < a[n - 1]. Now go from the top. When I am processing a[i], if 2^k - a[i] is still in the set, I match it. Otherwise, I continue.