In the first case, starting with p such that the highest power of 2 that divides p is an integer power of 2 (2^k for some integer k); then the highest power of 2 that divides p² is 2^2k; then the highest power of 2 that divides 2q² is also 2^2k; then the highest power of 2 that divides q is 2^(2k-1); therefore q must be a multiple of 2^(k-0.5); a noninteger power of 2.
This implies that there is a number 2^(0.5). It makes no claims as to whether or not this number is rational, or integer; it merely claims that such a number must exist. (Consider: if I had started instead with the equation x²-4=0, I would have ended up showing that a number of the form 4^(0.5) must exist - that number is rational, is indeed an integer).
Now, I think I can prove that an integer q which is a multiple of 2^(k-0.5) but which is not a multiple of 2^k, for integer k, does not exist; but I can only complete that proof by knowing in advance that 2^0.5 is irrational, so I can't use it to prove the irrationality of 2^0.5. I can easily prove that a rational number of the form 4^(k-0.5) for integer k does exist; indeed, an infinite number of such numbers exist (examples include 2, 8, 32).
No matter how forcefully that first passage conveys the irrationality of √2, it does not prove it.
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.
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