Let me expand my earlier comment to a partial answer of "why sparsity?". It is impossible for $|{\rm PSL}(2,p^{m})|$ to have four or fewer different prime factors when $p$ is an odd prime which is neither Fermat nor Mersenne and $m > 1$ is an integer. In fact, if $m >1$ is not a power of $2$, we will see that whenever $p >3$ is prime, then $|{\rm PSL}(2,p^{m})|$ has at least $5$ different prime divisors.
(These two facts "explain" why the non-squarefree $q$ appearing in your Python output are all either powers of $2$ or powers of Fermat or Mersenne primes, and why we only have $q$ of the form $p^{2n+1}$ ( for positive $n$) appearing for $p = 2$ or $p =3).$
If $m$ is itself even, then $p^{2m}-1$ is divisible by $p^{4}-1$. It is easy to check that $\frac{p-1}{2}, \frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are pairwise coprime. Also $\frac{p^{2}+1}{2}$ is odd, while one of $\frac{p \pm 1}{2}$ is even, and the other is odd. Hence if the odd prime $p$ is neither a Fermat prime nor a Mersenne prime, then $p^{4}-1$ is even, and has at least three different odd prime factors, so has $4$ or more different prime factors (none of which is $p$). Thus $|{\rm PSL}(2,p^{2n})|$ has $5$ or more different prime factors whenever $p$ is an odd prime which is neither Fermat nor Mersenne, and $n$ is any positive integer.
If $m$ is not a power of $2$, then $m$ is divisible by some odd prime $r$. We note below that that $p^{2r}-1$ has four or more prime factors whenever $r$ is an odd prime and $p$ is a prime greater than $3$.
For of the four integers $\frac{p-1}{2}, \frac{p+1}{2}, \frac{p^{r}-1}{p-1}$ and $\frac{p^{r}+1}{p+1},$ exactly one is even. Since $p >3$, the product $\frac{p+1}{2}\frac{p-1}{2}$ has at least two different prime factors, since $p-1$ and $p+1$ can't both be powers of $2$ for any prime $p$ greater than $3$.
Note also that Zsygmondy's Lemma tells us that there are primes $s$ and $t$ such that $p+s\mathbb{Z}$ has multiplicative order $r$ in the units of $(\mathbb{Z}/s\mathbb{Z})^{\times}$ and $p+t\mathbb{Z}$ has multiplicative order $2r$ in the units of $(\mathbb{Z}/t\mathbb{Z})^{\times}$. Then $s$ and $t$ are different (and both odd) and neither of them divides $p^{2}-1.$ Hence $p^{2r}-1$ has at least $4$ different prime divisors, and $|{\rm PSL}(2,p^{m})|$ has at least five prime divisors.