I apologize for answering my own question, but I think that the answer might be of some interest to others.
When I was trying to expand the cases where $a$ is a Fermat prime, I got the following theorem.
Theorem. For each odd prime $q$, there exist only finitely many pairs $(n,b)$ such that $\sigma(n)=q^b$.
Lemma 1. For every pairs of primes $(p,r)$, $\frac{p^{r^2}-1}{p-1}$ is not of the form $q^b$.
Proof. For $r=2$, supposing that $\frac{p^{2^2}-1}{p-1}=(p+1)(p^2+1)$ is of the form $q^b$ gives us that both $p+1$ and $p^2+1$ are of the form $q^b$. Since $\gcd(p+1,p^2+1)\le 2$, one has $q=2$. Since $p^2+1\ge 2^2+1=5$, one has $p^2+1\equiv 0\pmod 4$, which is impossible.
For $r\ge 3$, supposing that $\frac{p^{r^2}-1}{p-1}$ is of the form $q^b$ gives us $\frac{p^r-1}{p-1}$ is of the form $q^b$. Since $p^r\equiv 1\pmod q$, it follows that $\gcd\left(p^r-1,\frac{p^{r^2}-1}{p^r-1}\right)$ is a divisor of $r$. So, one has $r=q$. Setting $p^r=p^q=cq+1$ gives us
$$\frac{p^{r^2}-1}{p^r-1}=\frac{p^{q^2}-1}{p^q-1}\equiv \sum_{i=0}^{q-1}(cq+1)^i\equiv 1+\sum_{i=1}^{q-1}(1+icq)\equiv q+q^2\cdot\frac{c(q-1)}{2}\equiv q\pmod{q^2}.$$ Thus, one knows that $\frac{p^{r^2}-1}{p^r-1}$ can be divided by $q$ at most once. It follows that $\frac{p^{r^2}-1}{p^r-1}=q$. So, one has $q=\frac{p^{r^2}-1}{p^r-1}\gt \frac{p^r-1}{p-1}\ge q$, which is a contradiction. QED
Lemma 2. If $\frac{p^r-1}{p-1}$ is of the form $q^b$ for primes $p,r$ where $p\not=q$, then $r$ is a divisor of $q-1$.
Proof. Supposing that $r$ is not a divisor of $q-1$ gives us that there exist positive integers $m,n$ such that $rm-(q-1)n=1$. Since one has $p^r\equiv 1\pmod q$ and $p^{q-1}\equiv 1\pmod q$, one has $p^1\cdot p^{(q-1)n}\equiv p^{rm}\Rightarrow p\cdot (p^{q-1})^{n}\equiv (p^r)^m\Rightarrow p\equiv 1\pmod q$. Since $\gcd\left(p-1,\frac{p^r-1}{p-1}\right)$ is a divisor of $r$, one has $r=q$. Since $\frac{p^r-1}{p-1}\equiv q\pmod{q^2}$, one has $\frac{p^r-1}{p-1}=q$. Thus, it follows that $q=\frac{p^r-1}{p-1}\ge \frac{2^r-1}{2-1}\gt r=q$, which is a contradiction. QED
Theorem. For each odd prime $q$, there exist only finitely many pairs $(n,b)$ such that $\sigma(n)=q^b$.
Proof. By lemma 1, it is sufficient to prove that there exist only finitely many pairs of primes $(p,r)$ such that $\frac{p^r-1}{p-1}=q^b$.
Suppose that there exist infinitely many pairs of primes $(p,r)$ where $r\ge 3$ and $p\not=q$ such that $\frac{p^r-1}{p-1}=q^b$. Since $r$ is a divisor of $q-1$ by lemma 2, there exist only finitely many such $r$. So, for an $r$, there exist infinitely many primes $p$ such that $\frac{p^r-1}{p-1}=q^b$. Let $n=r-1\ge 2$. Let us separate $(p,r)$ into $n$ groups by the values of $m\pmod n$. For an $l\ (0\le l\le n)$, there exsit infinitely many pairs $(p,r)$ such that $\frac{p^r-1}{p-1}=q^m$ and $m\equiv l\pmod n$. Since $p^n\lt \frac{p^r-1}{p-1}\lt (p+1)^n$, one has $l\not=0$. Then, $$\frac{p^r-1}{p-1}=1+p+p^2+\cdots+p^n=q^{l+nk}=C_1x^n\ \ (C_1=q^l\ge q,x=q^c)$$ has infinitely many solutions $(p,x)$.
Setting $np=y-1,C_1n^n=C_2$ gives us
$$n^n+n^{n-1} (y-1)+⋯+n(y-1)^{n-1}+(y-1)^n=C_2 x^n.$$
So, there exist integers $a_0,a_1,\cdots,a_{n-1}$ such that
$$a_0+a_1y+a_2y^2+\cdots+a_{n-2}y^{n-2}=C_2x^n-y^n$$ where $a_{n-1}=0$.
Setting $C_3=\max\{|a_0|,|a_1|,\cdots,|a_{n-2}|\}$ gives us
$$C_2x^n-y^n\lt C_3(1+y+y^2+\cdots+y^{n-2})\lt C_3(y+1)^{n-2}\lt C_3(2y)^{n-2}.$$
So, setting $\frac xy=t,\frac{C_3\cdot 2^{n-2}}{C_2}$ gives us
$$t^n-\frac{1}{C_2}\lt\frac{C_4}{y^2}.$$
So, $$\left(\frac xy-\alpha_1\right)\left(\frac xy-\alpha_2\right)\cdots\left(\frac xy-\alpha_n\right)\lt\frac{C_4}{y^2}$$
has infinitely many solutions $(x_1,y_1),(x_2,y_2),\cdots,(x_k,y_k),\cdots$ where $|y_1|\lt|y_2|\lt\cdots$ and $\alpha_1,\alpha_2,\cdots,\alpha_n$ are distinct irrational solutions of $t^n-\frac{1}{C_2}=0$. Since $|y_k|\to\infty$ as $k\to\infty$, $$\left(\frac{x_k}{y_k}-\alpha_1\right)\left(\frac{x_k}{y_k}-\alpha_2\right)\cdots\left(\frac{x_k}{y_k}-\alpha_n\right)\to 0.$$
So, one may suppose $\left|\frac{x_k}{y_k}-\alpha_1\right|\to 0$. Let $d=\min|\alpha_{i_1}-\alpha_{i_2}|\ \ (i_1\not=i_2)$. Here, if one takes a sufficiently large $N$, one has $\left|\frac{x_k}{y_k}-\alpha_1\right|\lt\frac d2$ for $k\gt N$. One also has $\left|\frac{x_k}{y_k}-\alpha_i\right|\gt\frac d2\ \ (i\not=1)$. Thus, it follows that
$$\left|\frac{x_k}{y_k}-\alpha_1\right|\lt\frac{C_4}{{y_k}^2}\times\left(\frac 2d\right)^{n-1}=\frac{C_5}{{y_k}^2}$$where $C_5=C_4\left(\frac 2d\right)^{n-1}$.
Here, by the Roth-Ridout theorem, one has
$$\left|\frac{x_k}{y_k}-\alpha_1\right|\gt\frac{C_6}{{y_k}^{1.5}}$$
where $C_6$ is a constant independent of $x_k,y_k$. Thus, since one has $$\frac{C_6}{{y_k}^{1.5}}\lt \left|\frac{x_k}{y_k}-\alpha_1\right|\lt\frac{C_5}{{y_k}^2}\Rightarrow \frac{C_6}{{y_k}^{1.5}}\lt \frac{C_5}{{y_k}^2},$$
one has $y_k\lt C_7$ where $C_7$ is a constant independent of $x_k,y_k$. This is a contradiction. QED
