Aha. By editing my query a bit, I find that the number the Internet would have to count to is 2^(1.957×10^84).
That equals 10^(5.89×10^83), I think, which is on the order of 10^(10^84), which means that we'd need to count to a number that, expressed in decimal, would be 10^84 digits long. Ans we'd have to keep track of every single one of those digits, of course- if we could round it off like I just did, we wouldn't he here.
Not gonna happen anytime soon. That number is a far cry from 8, for sure.