Proving that no prime powers can be correct numbers I recently encountered the following problem:

A natural number $n$ is called correct if and only if the sum of the squares of $n$'s divisors equals $(n+3)^2$. Prove that no numbers of the form $n=p^m$, where $p$ is a prime and $m\in\Bbb{N}$ (prime powers) can be correct. (from the N. N. Mihaileanu Mathematics Contest – 8th Grade 2017, Romania)

First of all, I looked at the sum of the divisor squares:
$$S=\sum_{d\in D_{p^m}^+}d^2$$
Where $D_{p^m}^+$ is the set of $p^m$'s positive divisors. However, since $p^m$ is a prime power, this set can also be expressed as:
$$D_{p^m}^+=\{0\le k\le m\mid p^k\}$$
So it follows that our sum can simply be expressed as:
$$S=\sum_{k=0}^m (p^k)^2=\sum_{k=0}^m p^{2k}$$
Now, I tried to continue on with the proof using Reductio ad absurdum, by assuming that $S$ can equal $(n+3)^2$:
$$\sum_{k=0}^m p^{2k}=(p^m+3)^2$$
Subtracting $p^{2m}$ from both sides,
$$\sum_{k=0}^{m-1} p^{2k}=6p^m+9$$
Then, I tried evaluating the sum of that series, obtaining:
$$6p^m+9=\frac{p^{2m}-1}{p^2-1}$$

However, I am stuck here. I tried my best to avoid splitting in cases, but am currently unable to finish my proof. Any hints on what I should do next (preferably, without having to consider cases separately, but that works too)?
 A: Letting $x=p^m$, the key equation becomes $6x+9=(x^2-1)/(p^2-1)$, or
$$x^2-6(p^2-1)x-(9p^2-8)=0$$
for which the quadratic formula tells us
$$x=3(p^2-1)\pm\sqrt{9(p^2-1)^2+(9p^2-8)}=3(p^2-1)\pm\sqrt{9p^4-9p^2+1}$$
This implies the discriminant is a perfect square, which implies
$$36p^4-36p^2+4=(6p^2-3)^2-5$$
is a square. But the only two squares that differ by $5$ are $9$ and $4$, and
$$6p^2-3=3\implies p=\pm1$$
A: I think the simplest proof is a $\mod p$ argument (as suggested in one of the comments to the original question):
If $n=p^m$ then all of the divisors of $n$ apart from 1 are multiples of $p$, so
$\sum_{d|n} d^2 \equiv 1 \mod p$
and $(n+3)^2 = (p^m+3)^2 \equiv 9 \mod p$
If $1 \equiv 9 \mod p$ then $p=2$ and $p^m \in \{2,4,8,16\}$ - after 16, $\sum d^2 > (2^m+3)^2$. But we can eliminate $\{2,4,8,16\}$ by direct calculation. So
$\sum_{d|n} d^2 \ne (n+3)^2 \quad  \forall n=p^m$
A: I'll continue the argument from $$\sum_{k=0}^{m-1} p^{2k}=6p^m+9\tag{1}\label{eq1}$$
Firstly, assume $m=1$ or $m=2$. Then $\eqref{eq1}$ becomes $1=6p+9$ or $1+p^2=6p^2+9$, which is impossible. Thus $m\ge3$.
i) $p=2$. Then since $2^{2(m-1)}<6\times2^m+9<8\times2^m=2^{m+3}$, $m<5$. Manually checking the cases $m=3$ and $m=4$ gives no solutions.
ii) $p=3$, then $3^{2(m-1)}<6\times3^m+9<9\times3^m=3^{m+2}$, $m<4$. Manually checking the case $m=3$ gives no solution.
iii) $p=5$, then $5^{2(m-1)}<6\times5^m+9<25\times5^m=5^{m+2}$, $m<4$. Manually checking the case $m=3$ gives no solution.
iv) $p\ge7$, then $p^{2(m-1)}<6p^m+9<7p^m\le p^{m+1}$, $m<3$.
By i) to iv), we can conclude that no primes can be correct number.
Also, one can replace the arguments of ii) and iii) to modular arithmetic.
