Let $\sigma(x)$ be the sum of divisors of the positive integer $x$.
Here is my question:
When is $\sigma(q^{k-1})$ a square when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$?
MY ATTEMPT
$$r^2 = \frac{q^k - 1}{q - 1} = 1 + \ldots + q^{k-1} \equiv 1 + (k-1) \pmod 4 \equiv 1 \pmod 4$$ so that $r$ is odd. Additionally, $r \equiv 1 \pmod q$.
Alas, that is where I get stuck.
REVISED ATTEMPT
Let $q$ be a prime satisfying $q \equiv k \equiv 1 \pmod 4$.
I prove here that
$$\sigma(q^{k-1}) = s(q^k) \text{ is a square } \implies k = 1.$$
Proof
Assume to the contrary that $k > 1$. This implies that $k \geq 5$ (since $k \equiv 1 \pmod 4$).
Suppose that $$s(q^k) = s^2 = \sigma(q^k) - q^k = \sigma(q^{k-1}) = \frac{q^k - 1}{q - 1}.\tag{$*$}$$
$(*)$ implies that $(q-1)s^2 = q^k - 1$, which is equivalent to $$q(q^{k-1} - s^2) = q^k - qs^2 = 1 - s^2 = (1 + s)(1 - s) = -(s+1)(s-1).$$
Since $q$ is prime, we consider three cases:
Case 1: $q = s + 1$ $$\implies q - 1 = s \implies q^3 - 3q^2 + 3q - 1 = (q - 1)^3 = s^3 = (q - 1)s^2 = q^k - 1$$ $$\implies q^2 - 3q + 3 = q^{k-1} \geq q^4$$ This last inequality is a contradiction.
Case 2: $q = s - 1$ $$\implies q + 1 = s \implies q^{k-1} = s^2 - s - 1 = (q+1)^2 - (q+1) - 1$$ $$= q^2 + 2q + 1 - q - 1 - 1 = q^2 + q - 1$$ $$\implies q^2 + q - 1 = q^{k-1} \geq q^4$$ Again, this last inequality is a contradiction.
Case 3: $q = (s+1)(s-1)$ $$\implies q^{k-1} - s^2 = -1 \implies q^{k-1} = s^2 - 1 = q \implies k - 1 = 1 \implies k = 2.$$ However, we know that $k=2$ contradicts $k \equiv 1 \pmod 4$.
QED
In fact, more is true.
If $k=1$, then $s(q^k) \text{ is a square}$.
Therefore, we have the biconditional $$s(q^k) = \sigma(q^{k-1}) = \frac{q^k - 1}{q - 1}$$ is a square if and only if $k=1$.