It is $\sum_{n|d} \mu(n)( q^{d/n}-1)\gcd( (q^d-1)/(q^{d/n}-1), m)/\gcd(q^d-1,m)$

Because every m’th power that generates is an m’th power of a generator, it suffices to do the inclusion-exclusion for generators in the set of $m$’th powers, and this is the result.

The number of nonzero elements of a sub field that are $m$th powers is the number of elements, divided by the number of characters of order $m$, times the number of characters of order$m$ trivial on that sub field, explaining the terms in this formula.

$$| S \cap S^m | = \sum_{n|d} \mu(n) \frac{ \gcd ( m (q^{d/n}-1), q^{d}-1 )}{ \gcd (q^{d} - 1, m) } $$

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) } $.

Inclusion-exclusion gives our formula.