If $q$ is prime, can $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ be both squares when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$? 

This is related to this earlier MSE question.  In particular, it appears that there is already a proof for the equivalence
    $$\sigma(q^{k-1}) \text{ is a square } \iff k = 1.$$


Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
Here is my question:

If $q$ is prime, can $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ be both squares when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$?

MY ATTEMPT
Suppose that
$$\sigma(q^{k-1}) = a^2$$
and
$$\frac{\sigma(q^k)}{2} = b^2$$
for $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$.
Since $\sigma(q^k) = q^k + \sigma(q^{k-1})$, it follows that
$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) = q^k + a^2.$$
Additionally, congruence-wise we obtain
$$a^2 = \sigma(q^{k-1}) \equiv 1 + (k-1) \equiv k \equiv 1 \pmod 4,$$
from which it follows that $a$ is odd, and
$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) \equiv 1^1 + 1 \equiv 2 \pmod 4,$$
which implies that $b$ is likewise odd.
Now, using the definition of $\sigma(q^k)$ and $\sigma(q^{k-1})$ for $q$ prime, we derive
$$\frac{1}{2}\cdot\frac{q^{k+1} - 1}{q - 1} = b^2$$
and
$$\frac{q^k - 1}{q - 1} = a^2.$$
Assume to the contrary that
$$\frac{1}{2}\cdot\frac{q^{k+1} - 1}{q - 1} = b^2 \leq a^2 = \frac{q^k - 1}{q - 1}.$$
This assumption leads to
$$q^{k+1} - 1 \leq 2(q^k - 1)$$
which implies that
$$16 = {5^1}(5-2) + 1 \leq q^k(q - 2) + 1 = q^{k+1} - 2q^k + 1 \leq 0,$$
since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.  This results in the contradiction $16 \leq 0$.  Consequently, we conclude that $a < b$.
Furthermore, I know that
$$(q+1) = \sigma(q) \mid \sigma(q^k) = 2b^2$$
so that
$$\frac{q+1}{2} \leq b^2.$$
Finally, I also have
$$\frac{q^{k+1} - 1}{2b^2} = q - 1 = \frac{q^k - 1}{a^2}.$$
Alas, here is where I get stuck.
CONJECTURE (Open)

If $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$, then $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ are both squares when $k = 1$.

SUMMARY OF RESULTS SO FAR

zongxiangyi appears to have proven the implication
  $$\sigma(q^k)/2 \text{ is a square} \implies k = 1.$$

The proof of the following implication is trivial
  $$k = 1 \implies \sigma(q^{k-1}) \text{ is a square}.$$
  The truth value of the following implication is currently unknown:
  $$\sigma(q^{k-1}) \text{ is a square} \implies k = 1.$$
Together, the two results give
  $$\sigma(q^k)/2 \text{ is a square} \implies k = 1 \iff \sigma(q^{k-1}) \text{ is a square},$$
  so that $\sigma(q^{k-1})$ is a square if $\sigma(q^k)/2$ is a square.
Therefore, $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ are both squares (given $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$) when $\sigma(q^k)/2$ is a square.


Here are a couple of other approaches to consider which may be useful. First, your equation of
$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) = q^k + a^2 \tag{1}\label{eq1}$$
can be rewritten as
$$2b^2 - a^2 = q^k \tag{2}\label{eq2}$$
This is in the generalized Pell equation form of $x^2 - Dy^2 = N$. The blog Solving the generalized Pell equation explains how to solve this.
Next, note that
$$\sigma(q^{k-1}) = \sum_{i=0}^{k-1} q^i \tag{3}\label{eq3}$$
$$\sigma(q^{k}) = \sum_{i=0}^{k} q^i \tag{4}\label{eq4}$$
Thus, you can express $\sigma(q^{k})$ in terms of $\sigma(q^{k-1})$ as
$$\sigma(q^{k}) = q\sigma(q^{k-1}) + 1 \tag{5}\label{eq5}$$
As you stated, suppose
$$\sigma(q^{k-1}) = a^2 \tag{6}\label{eq6}$$
$$\frac{\sigma(q^k)}{2} = b^2 \iff \sigma(q^k) = 2b^2 \tag{7}\label{eq7}$$
Substituting \eqref{eq6} and \eqref{eq7} into \eqref{eq5} gives
$$2b^2 = qa^2 + 1 \iff 2b^2 - qa^2 = 1 \iff (2b)^2 - (2q)a^2 = 2 \tag{8}\label{eq8}$$
Wikipedia's Pell's equation page's Transformations section gives a related equation of
$$u^{2}-dv^{2}=\pm 2 \tag{9}\label{eq9}$$
and how it can be transformed into the Pell equation form of
$$(u^{2}\mp 1)^{2}-d(uv)^{2}=1 \tag{10}\label{eq10}$$
Here, $u = 2b$, $v = a$, $d = 2q$ and the right side of \eqref{eq8} is $2$, so \eqref{eq10} becomes
$$((2b)^2 - 1)^2 - (2q)(2ba)^2 = 1 \tag{11}\label{eq11}$$
This is in Pell's equation form of $x^2 - ny^2 = 1$. Since $n = 2q$ is not a perfect square, there are infinitely many integer solutions. However, among these solutions, you first need to check that $x$ is in the form $4b^2 - 1$, the determined $b$ divides $y = 2ba$ and then that $a$ and $b$ satisfy \eqref{eq6} and \eqref{eq7} for some $k \equiv 1 \pmod 4$.
As for your open conjecture, if $k = 1$, then isn't $\sigma(q^{k-1}) = \sigma(q^{0}) = 1$ and $\frac{\sigma(q^{k})}{2} = \frac{\sigma(q)}{2} = \frac{1 + q}{2}$, so having both of them be squares requires $q = 2b^2 - 1$ for some $b$ and, thus, is not always true for all primes $q \equiv 1 \mod 4$, e.g., for $q = 5$, you get $5 = 2b^2 - 1 \implies 6 = 2b^2 \implies b = \sqrt{3}$?
