But [p^2 contains exactly twice as many powers of 2 as p does] is only true because √2 is irrational, making the demonstration a circular proof.

Notice that you are claiming that all possible proofs of the statement "p^2 contains twice as many powers of 2 as p" require asserting without proof that sqrt(2) is irrational.

Why does the prime factorization of integers depend upon something that is, if not irrational, at least certainly not an integer? (Proof: 1^2 = 1, 2^2 = 4, and x <= x^2 by induction.)