Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. If $\sigma(y) < 2y$, $y$ is called deficient; if $\sigma(z) > 2z$, $z$ is called abundant.
Questions
(1) Can $\sigma(2^r)$ be abundant for some integer $r > 1$?
(2) If so, what conditions on $r$ guarantee abundance?
My Attempt
If $\sigma(2^r) = 2^{r + 1} - 1$ is a Mersenne prime, then it is deficient. Hence we consider the case when $\sigma(2^r) = 2^{r + 1} - 1$ is a composite Mersenne number.
I found this related question in MSE. It essentially states the following proposition:
Proposition A. If $n = {2^{m-1}}(2^m - 1)$, where $m$ is a positive integer such that $2^m - 1$ is composite, then $n$ is abundant.
Suppose to the contrary that $\sigma(2^r) = 2^{r+1} - 1$ is a deficient composite Mersenne number. Since $2^{r - 1}$ is also deficient, then subject to certain constraints on the abundancy index $$I\left(\sigma(2^r)\right)=\dfrac{\sigma\left(\sigma(2^r)\right)}{\sigma(2^r)},$$ (such as a suitable upper bound), then it might be possible to show that $$I(k) = I\left({2^{r-1}}\sigma(2^r)\right) < 2,$$ contradicting Proposition A.
"QED"
In Retrospect
However, I do realize that this argument is not rigorous, and may in fact be fallacious. This explains my second question.
Since integer $r > 1$ implies that $r - 1 \geq 1$, and since $$2 \mid 2^{r - 1}$$ and $$\gcd(2^{r - 1}, \sigma(2^r)) = 1,$$ then a condition on $r$ that guarantees abundance (by ensuring a contradiction against Proposition A) follows from the fact that the abundancy index is weakly multiplicative: $$\dfrac{3}{2} \cdot I\left(\sigma(2^r)\right) \leq I(2^{r-1})I\left(\sigma(2^r)\right) = I\left(2^{r-1}\sigma(2^r)\right) = I(k) < 2$$ so that $$I\left(\sigma(2^r)\right) < \dfrac{4}{3}.$$ This condition on $r$ will ensure a deficient $k = {2^{r-1}}\sigma(2^r)$ thereby contradicting Proposition A.
Consequently, the condition $$I\left(\sigma(2^r)\right) > \dfrac{4}{3}$$ is necessary for $\sigma(2^r)$ to be abundant. (Note that $$I\left(\sigma(2^r)\right) \neq \dfrac{4}{3}$$ since it is assumed that $r \neq 1$.)
Added June 2 2016