So there are 2^96 different private keys in the entire 2^256 space that can "unlock" for example this address: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN (Puzzle 64)? Can anyone confirm this?
Yes exactly and for a simple fact :
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN (and every other wallet address of this type) is encoded on 160 bits (and not on 256 bits like other parameter of bitcoin protocol)
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN is base58 encoded and is 3ee4133d991f52fdf6a25c9834e0745ac74248a4 in hexadecimal
raw public keys are encoded on 256 bits
a simple wallet address (p2pkh) is simply obtained by the function hash160(public_key)
So if you have the possibility to browse the entire 1-2^256 space and to compute the hash160 function for every hash160(public_key) derived from 1-2^256 private keys you will find an average of
2^(256-160) = 2^96 public key with hash160=16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN .
But even if 2^96 seems big its far away smaller from 2^256. And it's pretty impossible to have a collision between two random public key in the using age of bitcoin.