$$1 + \frac{1}{p} \le \frac{\sigma(p^l)}{p^l} = \frac{p - \frac{1}{p^l}}{p-1} < \frac{p}{p-1},$$

$$(2^{k+1} - 1) \left(1 + \frac{1}{p}\right) \le \frac{\sigma(n)}{p^l} = 2^{k+1} - \frac{2}{p^l} < (2^{k+1} - 1) \frac{p}{p-1}.$$
By the first inequality, $(2^{k+1}-1)(1+1/p) < 2^{k+1}$, so $p > 2^{k+1}-1$, i.e., $p \ge 2^{k+1}+1$ since $p$ is odd. On the other hand, $p < 2^{k+1}+2(p-1)/p^l$, by the second inequality, so $2(p-1) > p^l$, and consequently $l=1$ and $p = 2^{k+1}+1$.
To rule out the case $n = p^k q^l$, where $p$ and $q$ are distinct odd primes, and $k$ and $l$ are positive integers, write
$$2 - \frac{2}{n} = \frac{\sigma(n)}{n} = \frac{\sigma(p^k)}{p^k} \cdot \frac{\sigma(q^l)}{q^l} < \frac{p}{p-1} \cdot \frac{q}{q-1}.$$
Alternatively, but equivalently,
$$\frac{1}{p-1} + \frac{1}{q-1} + \frac{1}{(p-1)(q-1)} + \frac{2}{n} > 1,$$
so $\min(p, q) = 3$, say $p=3$. Then $3/(q-1)+4/n > 1$, and it follows that $q=5$ and $k=l=1$, i.e., $n=15$ which does not satisfy the condition $\sigma(n) = 2n-2$. This completes the proof.