I'm reading Richard Ryan's article "A simpler dense proof regarding the abundancy index" and got stuck in his proof for Theorem 2. The Theorem is stated as follows:

Suppose we have a fraction of the form $\frac{2n-1}{n}$, where $2n-1$ is prime.

(i) ...

(ii) If $n$ is odd and $I(b)=\frac{2n-1}{n}$ for some $b$, then $b$ is odd; moreover, if $2n-1$ does not divide $b$, then $b(2n-1)$ is a perfect number.

Ryan's proof: Suppose that $n$ is odd and $b$ is even. Let $m$ be the greatest integer such that $2^m$ divides $b$. Once again, there is a prime factor, $q$, of $\sigma(2^m)$ that divides $b$. Thus $I(b) > I(2^mq) > 2$ and we have a contradiction. Finally, if the prime number $2n - 1$ does not divide $b$ then, since $I$ is multiplicative, $I(b(2n - 1)) = 2$.

The 'once again' comes from (i): ... let $I(b)=\frac{2n-1}{n}$ for even $b$ and $m$ be the greatest integer such that $2^m$ divides $b$. There is a prime factor, $q$, of $\sigma(2^m)$ that also divides $b$ since $\sigma(2^m) = 2^{m+1} - 1 \neq 2n - 1$. Thus

$I(b) > I(2^mq) \geq \frac{2^{m+1}-1}{2^m}\cdot\frac{2^{m+1}}{2^{m+1}-1}=2$,

which contradicts $I(b)=\frac{2n-1}{n}$.

I see that $\sigma(2^m)=2^{m+1}-1\neq 2n-1$, because $n$ is odd and thus $2n$ is not a power of two. Ryan writes that $\sigma(2^m)$ must have the prime factor $q$ and arrives at a contradiction through that. Could someone explain this to me? I'm completely lost with the bit from (i).

Abundancy index $I$ is defined as $I(n)=\frac{\sigma(n)}{n}$, where $\sigma$ counts the sum of divisors. Both $I$ and $\sigma$ are multiplicative.

I did not find a tag for abundancy index.