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You have the logic backwards. If there can be non-isomorphic groups with a structure-preserving bijection between their subgroups, certainly there can be non-isomorphic groups with the same number of subgroups.
There are many many adjectives which describe trees. I'd recommend you spend some time with a nice survey article (or maybe even the wikipedia page) to get acquainted with them. Your question about how balanced a tree is sounds potentially interesting, but I think you'll be able to ask it better after having done some more reading.
+1: Thanks for this interpretation -- I think this makes a lot of sense, though I'm going to spend some time working it out more carefully. Do you have a reference for the historical comment? I'd like to understand the historical development better.
So certainly a lot of these techniques become hard/impossible as you move away from quadratic Diohpantine equations, right? I'm thinking methods 1, 2, and 7 in particular -- and 3 gets harder and harder as well. Maybe this should just be an exercise for myself, but it is clear to you that it would be rather difficult to find examples where the Hilbert Class Field approach would the best/only one?
@Agol: Well, perhaps not directly, but it sounds fascinating regardless, so +1. If there's an answer there to be elaborated on, I'd love to see it. This will save me from asking the new question "What was Agol talking about when he said..." :)
@Dev Sinha: Okay, fair point. On the other hand, there are a lot of important statements in group cohomology which involve trivial coefficients: The Schur multiplier (or Hopf's Integral Homology Formula) comes to mind, as does the interpretation of the relation rank of a pro-$p$-group $G$ as $\operatorname{dim}H^2(G,\mathbb{F}_p)$.