The key to this sequence, and you may have spotted it in the first few terms I gave in the question, are the Fermat numbers: F_n = 2^2^n +1. The first few Fermat numbers are 3, 5, 17, 257, 65537. All five of those numbers are prime, so the totient function of any of them, or any product of a combination of them is as follows:

phi(F_a x F_b x F_c) = 2^2^a x 2^2^b x 2^2^c = 2^(2^a + 2^b + 2^c).

These first five values cover the subscripts 0 to 4, so by using the properties of binary numbers, we can produce an odd number with a totient function of the for 2^k for any k that can be made by combining different powers of 2.

To demonstrate, a selected few are as follows:

1 = 2^0

2 = 2^1

3 = 2^1 + 2^0

…

7 = 2^2 + 2^1 + 2^0

8 = 2^3

…

30 = 2^4 + 2^3 + 2^2 + 2^1

31 = 2^4 + 2^3 + 2^2 + 2^1 + 2^0

For example, for k=7, the totient function of the number F_2 x F_1 x F_0 = 17 x 5 x 3 = 255 is 2^(2^2 + 2^1 + 2^0) = 2^(4+2+1) = 2^7.

So this can give us any power of 2 as a totient of an odd number, as long as the Fermat numbers are prime, however as suggested in the question, this doesn’t continue forever. This is because the next Fermat number F_5 = 4294967297 = 641 x 6700417 and as such is not prime. So when we try to use this method to reach 2^32, the totient function of 4294967297 is not the desired 2^32 = 4294967296, but is in fact 640 x 6700416 = 4288266240. In fact because there are no more known Fermat primes, beyond the first 31 there are no more powers of 2 known to be the totient of an odd number.