This is a conjucture that I created :
Let $\,n = (2k+1)^2 \,\, $with $k\in \mathbb{N}$ and so $n>1$, and let $$\,\,A = \sum_{d \in \mathbb{N}; \ d|n} d.$$ Then $n^m$ is never divisible by $A$ for every $m \in \mathbb{N}$ .
I found a proof for the simpler case with $n$ odd but not a perfect square :
An odd number which is not a square has a even number of divisors all odd . So their sum is even but the number raised to the $m$-th power is odd.
So if the conjucture was true then the theorem would be true for all odd numbers greater than $1$.
However I have no idea how to proceed to prove it in the case of an odd perfect square.
It seems rather linked with perfect numbers.
Here is a heuristic argument (that is too long for a comment) for why you would not expect $A$ to divide $n^3$, or any $n^k$. If we consider the argument for $k$ arbitrary, what we are asking is: for odd $n > 1$, is it always the case that $\sigma(n)$ has a prime factor that is not a factor of $n$? As you've pointed out, if $n$ is non-square, then $\sigma(n)$ is even, so the conjecture is true, and this leaves the case when $n$ is a square.
If $n = p_1^{2k_1}p_2^{2k_2} \cdots p_l^{2k_l}$, then $$ \sigma(n) = \prod_{i = 1}^l \sigma(p_i^{2k_i}) = \prod_{i=1}^l \frac{p_i^{2k_i + 1} - 1}{p_i - 1}. $$ So, finding a counterexample comes down to the following problem: find a set of odd primes $\{p_1, \ldots, p_l\}$ with exponents $k_1, \ldots, k_l$ such that for each $i$, the number $$ \frac{p_i^{2k_i + 1} - 1}{p_i - 1} $$ factorizes into the primes $p_1, \ldots, p_l$. Now this number grows very fast (in $k$), and so it is unlikely to accidentally hit a fairly smooth number. Among the first 100 odd primes $p_i$, with $k_i$ ranging up to 10, there are only a few situations where this number is even $p_i$-smooth -- a necessity for the largest prime dividing $n$. This already shows that $n$ must be quite large, and this is in a sense a very weak argument.