Enumeration of field generators of a finite field over $\mathbb{F}_{q}$, which are $m^{th}$ powers in the same field

Consider the finite field extension $$\mathbb{F}_{{q}^{d}}$$ over $$\mathbb{F}_{q}$$, where $$q=p^a$$ for some prime $$p$$. We assume $$d\geq 2$$. Let, $$S=\{ \alpha \in \mathbb{F}_{q^d}\hspace{0.1 cm} | \hspace{0.1 cm} \mathbb{F}_{q}(\alpha)=\mathbb{F}_{q^d} \}$$

In other words, $$S$$ consists of all those elements in $$\mathbb{F}_{q^d}$$, whose minimal polynomial over $$\mathbb{F}_{q}$$ has degree $$d$$, or to say another way $$S$$ consists of all field generators of $$\mathbb{F}_{q^d}$$ Over $$\mathbb{F}_{q}$$. Let, $$S^m= \{ s^m | s\in S\}$$, where $$m$$ is a positive integer $$\geq 2$$. Then, $$|S\cap S^m|=?$$

More, precisely, $$S\cap S^m$$ is the set of all field generators in $$\mathbb{F}_{q^d}$$ (over $$\mathbb{F}_q$$), which are $$m^{th}$$ powers in $$\mathbb{F}_{q^d}$$.

I calculated, this for $$m=2$$. The answer depends on whether $$d$$ is odd or even. We have,

$$|S\cap S^2|= \begin{cases} \frac{|S|}{2} & if \hspace{0.2 cm} d \text{ is odd}\\ \frac{1}{2}[|S|-\frac{(q^{d/2}-1)}{d}] & if \hspace{0.2 cm} d \text{ is even} \end{cases}$$

And, $$|S|=dM(d,q)$$, where $$M(d,q)$$ denote the number of irreducible polyomials of degree $$d$$ over $$\mathbb{F}_{q}$$.

I couldn't generalize my method to $$m>2$$. My, idea is that this problem seems to have been well studied in the literature of finite fields. So, I am hoping for some kind of help or suitable references in this case.

Thank you!

$$| S \cap S^m | = \sum_{n|d} \mu(n) \frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) }$$
First note that $$|S \cap S^m | = |S \cap \mathbb F_{q^d}^m |$$ because every $$m$$th power that generates is an $$m$$th power of a generator.
We can count elements of $$S$$ by an inclusion-exclusion argument, subtracting and adding the number of elements in subfields. This gives the term $$\mu(n) q^{d/n}$$, or $$\mu(n)( q^{d/n}-1)$$ if we only count nonzero elements. To count elements of $$S$$ that are $$m$$th powers, we use inclusion-exclusion to count the number of $$m$$th powers in subfields.
To count the number of elements of $$\mathbb F_{q^{d/n}}^\times$$ that are $$m$$th powers in $$\mathbb F_{q^d}^\times$$, we observe that their $$m$$th roots are both $$m (q^{d/n}-1)$$st roots of unity and $$(q^d-1)$$st roots of unity, hence are $$\gcd ( m (q^{d/n}-1),(q^d-1))$$th roots of unity, and each of them has $$\gcd(q^d-1, m)$$ $$m$$th roots in $$\mathbb F_{q^d}^\times$$, so the total number of them is $$\frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) }$$.