I cannot answer your question, but I proved two claims which might be useful to answer your question.
This answer proves the following claims :
Claim 1 : In $\text{mod}\ 4$, $$\sigma(m^2)\equiv 1\iff \frac{m^2}{\gcd(m^2, \sigma(m^2))}\equiv \frac{\sigma(m^2)}{\gcd(m^2, \sigma(m^2))}\iff \text{$\dfrac{2p^k-\sigma(p^k)}{4}\ $ is even}$$
Claim 2 : $$\sigma(m^2)\equiv \frac{(t+1)t}{8}\bigg(4+(t^2+1)(v-1)\bigg)\pmod 8$$
where $p\equiv t\pmod{16}$ and $k\equiv v\pmod{16}$.
(Though Claim 2 is "weaker" than what you've already known, it may be useful since all the cases are included in one expression.)
Claim 1 : In $\text{mod}\ 4$, $$\sigma(m^2)\equiv 1\iff \frac{m^2}{\gcd(m^2, \sigma(m^2))}\equiv \frac{\sigma(m^2)}{\gcd(m^2, \sigma(m^2))}\iff \text{$\dfrac{2p^k-\sigma(p^k)}{4}\ $ is even}$$
Proof :
In this proof, we consider in $\text{mod}\ 4$.
Let $$G= \gcd(m^2, \sigma(m^2)),\quad m^2=Ga,\quad\sigma(m^2)=Gb$$
where $a,b$ are odd such that $\gcd(a,b)=1$.
Then, we see that the equation A is equivalent to
$$2D(m^2)s(m^2) = G^2D(p^k)s(p^k)$$
i.e.
$$2(2Ga-Gb)(Gb-Ga)=G^2\bigg(2p^k-\sigma(p^k)\bigg)\bigg(\sigma(p^k)-p^k\bigg)$$
i.e.
$$2(2a-b)(b-a)=\bigg(\underbrace{2p^k-\sigma(p^k)}_{\equiv\ 0}\bigg)\bigg(\underbrace{\sigma(p^k)-p^k}_{\equiv\ 1}\bigg)$$
i.e
$$(2a-b)(b-a)=2c(4d+1)\tag1$$
where $c,d$ are integers such that
$$2p^k-\sigma(p^k)=4c,\qquad \sigma(p^k)-p^k=4d+1$$
So, we have, from $(1)$,
$$3ab-2a^2-b^2\equiv 2c\tag2$$
Since $a^2\equiv b^2\equiv 1$, we finally have, from $(2)$,
$$ab\equiv 2c+1\tag3$$
Now, from $(3)$, we get the followings :
If $G\equiv a\equiv 1$ and $b\equiv 1$, then $\sigma(m^2)\equiv 1$, and $c$ is even.
If $G\equiv a\equiv 1$ and $b\equiv 3$, then $\sigma(m^2)\equiv 3$, and $c$ is odd.
If $G\equiv a\equiv 3$ and $b\equiv 1$, then $\sigma(m^2)\equiv 3$, and $c$ is odd.
If $G\equiv a\equiv 3$ and $b\equiv 3$, then $\sigma(m^2)\equiv 1$, and $c$ is even.
Therefore, we can say that
$$\sigma(m^2)\equiv 1\iff a\equiv b\iff \text{$c$ is even.}\quad\blacksquare$$
Claim 2 : $$\sigma(m^2)\equiv \frac{(t+1)t}{8}\bigg(4+(t^2+1)(v-1)\bigg)\pmod 8$$
where $p\equiv t\pmod{16}$ and $k\equiv v\pmod{16}$.
Proof :
We have
$$\sigma(m^2)\equiv p^km^2\bigg(\frac{\sigma(p^k)}{2}\bigg)^{-1}\equiv p^k\bigg(\frac{\sigma(p^k)}{2}\bigg)^{-1}\equiv p^k\bigg(\frac{\sigma(p^k)}{2}\bigg)\pmod 8$$
Let $p=16s+t\ (t>1), k=16u+v$ where $t^4\equiv 1\pmod{16}$.
Then, there is an integer $z$ such that
$$\begin{align}\frac{\sigma(p^k)}{2}&\equiv \frac 12(1+t+\cdots +t^k)\pmod{8}
\\\\&\equiv \frac{t^{k+1}-1}{2(t-1)}\pmod{8}
\\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2(t^4)^{\frac{k-1}{4}}\bigg)\pmod{8}
\\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2(t^4-1+1)^{\frac{k-1}{4}}\bigg)\pmod{8}
\\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2\sum_{j=0}^{\frac{k-1}{4}}\binom{\frac{k-1}{4}}{j}(t^4-1)^j\bigg)\pmod{8}
\\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2\bigg(1+\frac{k-1}{4}(t^4-1)+(t^4-1)^2z\bigg)\bigg)\pmod{8}
\\\\&\equiv \frac{1}{2(t-1)}\bigg(-1+t^2+\frac{k-1}{4}(t^4-1)t^2\bigg)\pmod{8}
\\\\&\equiv \frac{t+1}{2}+\frac{k-1}{2}t^2\frac{(t^2+1)(t+1)}{4}\pmod{8}
\\\\&\equiv \frac{t+1}{2}+\frac{v-1}{2}t^2\frac{(t^2+1)(t+1)}{4}\pmod{8}\end{align}$$
Also, since $t^2\equiv 1\pmod 8$,
$$p^k\equiv (16s+t)^{16u+v}\equiv t^v(t^4)^{4u}\equiv t^v\equiv t\pmod 8$$
Therefore, we finally get
$$\sigma(m^2)\equiv \frac{(t+1)t}{8}\bigg(4+(t^2+1)(v-1)\bigg)\pmod 8$$
This holds even for $t=1$. $\ \blacksquare$
Examples :
If $p\equiv k\equiv 1\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.
If $p\equiv 5,k\equiv 13\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.
If $p\equiv k\equiv 9\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.
If $p\equiv 13,k\equiv 5\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$.
Therefore, we can say that if $p+k\equiv 2\pmod{16}$, then $\sigma(m^2)\equiv 1\pmod 8$. (This is what you've already written in your answer.)
