The paragraph, of course, was talking about integer powers of 2 that divide p. As in, the largest number 2^k such that 2^k divides p and k is an integer.
The largest real power of 2 that divides p is, of course, p itself, as 2^log_2(p) = p.
Looking over my post again, after a good night's sleep, I see that it wasn't as coherent as it appeared to me yesterday. Let me see if I can put my point a little more clearly.
The paragraph centers its claim of the irrationality of √2 on the idea that p² contains exactly twice as many powers of 2 as p does. But that is only true because √2 is irrational, making the demonstration a circular proof.
Consider. If √2 were rational, in the form of z/y for some coprime integers z and y, then it would be easy to find an integer that is not itself an integer power of 2, but whose square is an integer power of 2; z would be such a number.
Here's the new thread for posting quotes, with the usual rules: