If $k$ is arbitrarily large in the equations

$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$

then

$$\lim_{k \to \infty}{\frac{\sigma(n^2)}{q^k}}=\lim_{k \to \infty}{\frac{2n^2}{\sigma(q^k)}}=\lim_{k \to \infty}{\bigg(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\bigg)}=0$$

while

$$\gcd(n^2,\sigma(n^2))>1$$

for a fixed Euler part $n^2$. This implies that $k$ is bounded.

**UPDATE (June 11, 2017)****This claim is flawed. Unconditionally, we know that $q^k < n^2$. This implies that $k$ and $n$ are dependent.**