From the comments:

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha}))={\begin{cases}{\dfrac{N}{q^{\alpha} \sigma(q^{\alpha})}, \alpha \equiv 0 \pmod 2 \\ \dfrac{2N}{q^{\alpha} \sigma(q^{\alpha})}, \alpha \equiv 1 \pmod 2}\end{cases}}.$$

Since it is known that the index $$\sigma(N/q^{\alpha})/q^{\alpha} \geq 3$$ and since $$\sigma(q^{\alpha})\sigma(N/q^{\alpha})=\sigma(N)=2N,$$ because $N$ is perfect and $\gcd(q^{\alpha},N/q^{\alpha})=1$, then we obtain $$\sigma(N/q^{\alpha})/q^{\alpha} = \frac{2N}{q^{\alpha} \sigma(q^{\alpha})}.$$

It follows that

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha})) \geq \dfrac{3}{2} > 1,$$

which proves the original claim.

From the comments:

$$\gcd(N/q^\alpha,\sigma(N/q^\alpha)) = \begin{cases} \dfrac{N}{q^\alpha \sigma(q^\alpha)}, & \alpha \equiv 0 \pmod 2, \\[8pt] \dfrac{2N}{q^\alpha \sigma(q^\alpha)}, & \alpha \equiv 1 \pmod 2. \end{cases}$$

Since it is known that the index $$\sigma(N/q^{\alpha})/q^{\alpha} \geq 3$$ and since $$\sigma(q^{\alpha})\sigma(N/q^{\alpha})=\sigma(N)=2N,$$ because $N$ is perfect and $\gcd(q^{\alpha},N/q^{\alpha})=1$, then we obtain $$\sigma(N/q^{\alpha})/q^{\alpha} = \frac{2N}{q^{\alpha} \sigma(q^{\alpha})}.$$

It follows that

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha})) \geq \dfrac{3}{2} > 1,$$

which proves the original claim.

From the comments:

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha}))={\begin{cases}{\dfrac{N}{q^{\alpha} \sigma(q^{\alpha})}, \alpha \equiv 0 \pmod 2 \\ \dfrac{2N}{q^{\alpha} \sigma(q^{\alpha})}, \alpha \equiv 1 \pmod 2}\end{cases}}.$$

Since it is known that the index $$\sigma(N/q^{\alpha})/q^{\alpha} \geq 3$$ and since $$\sigma(q^{\alpha})\sigma(N/q^{\alpha})=\sigma(N)=2N,$$ because $N$ is perfect and $\gcd(q^{\alpha},N/q^{\alpha})=1$, then we obtain $$\sigma(N/q^{\alpha})/q^{\alpha} = \frac{2N}{q^{\alpha} \sigma(q^{\alpha})}.$$

It follows that

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha})) \geq \dfrac{3}{2} > 1,$$ which proves the original claim.

From the comments:

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha}))={\begin{cases}{\dfrac{N}{q^{\alpha} \sigma(q^{\alpha})}, \alpha \equiv 0 \pmod 2 \\ \dfrac{2N}{q^{\alpha} \sigma(q^{\alpha})}, \alpha \equiv 1 \pmod 2}\end{cases}}.$$

Since it is known that the index $$\sigma(N/q^{\alpha})/q^{\alpha} \geq 3$$ and since $$\sigma(q^{\alpha})\sigma(N/q^{\alpha})=\sigma(N)=2N,$$ because $N$ is perfect and $\gcd(q^{\alpha},N/q^{\alpha})=1$, then we obtain $$\sigma(N/q^{\alpha})/q^{\alpha} = \frac{2N}{q^{\alpha} \sigma(q^{\alpha})}.$$

It follows that

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha})) \geq \dfrac{3}{2} > 1,$$

which proves the original claim.