When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$? What are the exceptions?

This post is taken verbatim from this MSE question.

(This question is related to this earlier one.)

Let $\sigma(x)$ be the sum of divisors of the positive integer $x$.  The greatest common divisor of the integers $a$ and $b$ is denoted by $\gcd(a,b)$.

Here are my questions:

When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$?  What are the exceptions?

I tried searching for examples and counterexamples via Sage Cell Server, it gave me these outputs for the following GP scripts:

    for(x=1, 100, if(gcd(x,sigma(x^2))==gcd(x^2,sigma(x^2)),print(x)))

All positive integers from $1$ to $100$ (except for the integer $99$) satisfy $\gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))$.

    for(x=1, 1000, if(gcd(x,sigma(x^2))<>gcd(x^2,sigma(x^2)),print(x)))

The following integers in the range $1 \leq m \leq 1000$ DO NOT satisfy $\gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))$:

$$99 = {3^2}\cdot{11}$$

$$154 = 2\cdot 7\cdot 11$$

$$198 = 2\cdot{3^2}\cdot{11}$$

$$273 = 3\cdot 7\cdot 13$$

$$322 = 2\cdot 7\cdot 23$$

$$396 = {2^2}\cdot{3^2}\cdot{11}$$

$$399 = 3\cdot 7\cdot 19$$

$$462 = 2\cdot 3\cdot 7\cdot 11$$

$$469 = 7\cdot 67$$

$$495 = {3^2}\cdot 5\cdot 11$$

$$518 = 2\cdot 7\cdot 37$$

$$546 = 2\cdot 3\cdot 7\cdot 13$$

$$553 = 7\cdot 79$$

$$620 = {2^2}\cdot 5\cdot 31$$

$$651 = 3\cdot 7\cdot 31$$

$$693 = {3^2}\cdot 7\cdot 11$$

$$741 = 3\cdot 13\cdot 19$$

$$742 = 2\cdot 7\cdot 53$$

$$770 = 2\cdot 5\cdot 7\cdot 11$$

$$777 = 3\cdot 7\cdot 37$$

$$792 = {2^3}\cdot{3^2}\cdot 11$$

$$798 = 2\cdot 3\cdot 7\cdot 19$$

$$903 = 3\cdot 7\cdot 43$$

$$938 = 2\cdot 7\cdot 67$$

$$966 = 2\cdot 3\cdot 7\cdot 23$$

$$990 = 2\cdot{3^2}\cdot 5\cdot 11$$

MY ATTEMPT

I know that primes $m_1 := p$ and prime powers $m_2 := q^k$ satisfy the equation, since then we have

$$\gcd(m_1, \sigma({m_1}^2)) = \gcd(p, \sigma(p^2)) = 1 = \gcd(p^2, \sigma(p^2)) = \gcd({m_1}^2, \sigma({m_1}^2)),$$

and

$$\gcd(m_2, \sigma({m_2}^2)) = \gcd(q^k, \sigma(q^{2k})) = 1 = \gcd(q^{2k}, \sigma(q^{2k})) = \gcd({m_2}^2, \sigma({m_2}^2)).$$

This shows that there are infinitely many solutions to the equation

$$\gcd(m, \sigma(m^2)) = \gcd(m^2, \sigma(m^2)).$$

Follow-Up Questions

What can be said about solutions to $\gcd(m, \sigma(m^2)) = \gcd(m^2, \sigma(m^2))$ for which the number of distinct prime factors $\omega(m)$ satisfies

(a) $\omega(m)=2?$

(b) $\omega(m)=3?$