Cyclic subgroups of $\operatorname{GL}(n,q)$ of order $q^n - 1$ are called Singer cyclic subgroups. The following statement seems to be well-known:
Any two Singer cyclic subgroups of $\operatorname{GL}(n,q)$ are conjugate.
I don't know how to prove this in general. Sometimes, $q^n - 1$ is prime such that Sylow can be used (Example: $q=2, n=3$).
Some more background:
Elements of order $q^n - 1$ (and thus Singer cyclic subgroups) always exist: Let $\alpha$ be a primitive element of $\operatorname{GF}(q^n)$ and look at the $\operatorname{GF}(q)$-linear map $\operatorname{GF}(q^n) \to \operatorname{GF}(q^n)$ given by $x \mapsto \alpha x$.
$q^n - 1$ is the largest element order in $\operatorname{GL}(n,q)$, as by Cayley-Hamilton, the $\operatorname{GF}(q)$ vector space spanned by $I,A,A^2,A^3,\ldots$ has at most dimension $n$.
The elements of order $q^n - 1$ are not necessarily conjugate. For example, in $\operatorname{GL}(3,2)$ there are two conjugacy classes of elements of order $7$. However, the generated Singer cyclic subgroups are conjugate.
Here is a slightly expanded version of the proof mentioned by Dietrick Burde using the representation theory of cyclic groups.
A cyclic subgroup of order $q^n-1$ clearly clearly acts irreducibly on ${\mathbb F}_q^n$, so the minimal polynomial of a generator $g$ is a factor of $x^{q^n-1}-1$ of degree $n$ that is irreducible over ${\mathbb F}_q$.
So the subalgebra of $A_g \le M_n(q)$ generated by $g$ is isomorphic to the field of order $q^n$.
Since there is a unique such field up to isomorphism, for any other $g' \in {\rm GL}(n,q)$ of order $q^n-1$ generating a matrix algebra $A_{g'}$, there is an ${\mathbb F}$-isomorphism $\phi:A_g \to A_{g'}$, and the matrix defined by $\phi$ conjugates the set $\langle g \rangle = A_g \setminus \{ 0 \}$ to $\langle g' \rangle$.