Here are some partial results. I hope that they will help narrow down the search for counterexamples. Suppose that $pq^2=n|\sigma_2(n)=1+p^2+q^2+p^2q^2+q^4$.

Claim 1: We must have $q\equiv 3\pmod 4$.
Proof: Suppose $q\equiv 1\pmod 4$. Then $q$ remains prime/irreducible in $\mathbb{Z}[i]$. So $q| (1 + p^2) = (1+ip)(1-ip)\Rightarrow q$ divides one of $1\pm ip$ in $\mathbb{Z}[i]$, which is impossible.

Claim 2: We must have $p\equiv1\pmod 3$.
Proof: We have $p|(1+q^2+q^4) = (1+q^2)^2 - q^2 = (1+q+q^2)(1-q+q^2)$, so $p$ divides one of these factors. Suppose that $p$ divides $1+q+q^2$. Then $\mathbb{F}_p$ contains the cube roots of unity, so $p\equiv1\pmod 3$. Similarly, if $p$ divides the second factor, then $\mathbb{F}_p$ contains the sixth roots of unity.

Note that for each such $p$, the last observation implies that there are at most two choices for $q$, for which the required divisibility can possibly hold. Namely, if $p\equiv1\pmod 3$, then there exist integers $A$ and $B$ such that $p=A^2 + 27B^2$ (this is due to Gauss) and the non-trivial cube roots of 1 mod $p$ are $$ \frac{A+9B}{A-9B}\text{ and }\frac{A-9B}{A+9B}. $$ Since $q$ is a cube root of 1 mod $p$ and since we also know that $q^2|(1+p^2)$, we have that $q$ is less than $p$ and so it must be one of the two numbers between 3 and $p-2$ satisfying the above congruences. Of course, often, these numbers won't be congruent to 3 mod 4 or won't be prime. Whether that's never the case, I am not sure.

Here are some partial results. I hope that they will help narrow down the search for counterexamples. Suppose that $pq^2=n|\sigma_2(n)=1+p^2+q^2+p^2q^2+q^4$.

Claim 1: We must have $q\equiv 1\pmod 4$.
Proof: Suppose $q\equiv 3\pmod 4$. Then $q$ remains prime/irreducible in $\mathbb{Z}[i]$. So $q| (1 + p^2) = (1+ip)(1-ip)\Rightarrow q$ divides one of $1\pm ip$ in $\mathbb{Z}[i]$, which is impossible.

Claim 2: We must have $p\equiv1\pmod 3$.
Proof: We have $p|(1+q^2+q^4) = (1+q^2)^2 - q^2 = (1+q+q^2)(1-q+q^2)$, so $p$ divides one of these factors. Suppose that $p$ divides $1+q+q^2$. Then $\mathbb{F}_p$ contains the cube roots of unity, so $p\equiv1\pmod 3$. Similarly, if $p$ divides the second factor, then $\mathbb{F}_p$ contains the sixth roots of unity.

Note that for each such $p$, the last observation implies that there are at most two choices for $q$, for which the required divisibility can possibly hold. Namely, if $p\equiv1\pmod 3$, then there exist integers $A$ and $B$ such that $p=A^2 + 27B^2$ (this is due to Gauss) and the non-trivial cube roots of 1 mod $p$ are $$ \frac{A+9B}{A-9B}\text{ and }\frac{A-9B}{A+9B}. $$ Since $q$ is a cube root of 1 mod $p$ and since we also know that $q^2|(1+p^2)$, we have that $q$ is less than $p$ and so it must be one of the two numbers between 3 and $p-2$ satisfying the above congruences. Of course, often, these numbers won't be congruent to 1 mod 4 or won't be prime. Whether that's never the case, I am not sure.