Let $\gamma$ be a primitive element of $\mathbb{F}_q$.
How to prove that polynomial $x^{q-1}-\gamma$ is irreducible over $\mathbb{F}_q$?
Denote $\alpha$ as a root of $x^{q-1}-\gamma$, I want to show that $[\mathbb{F}_q(\alpha):\mathbb{F}_q]$is $q-1$. One can see $\alpha$ has order $(q-1)^2$, but how does this related to degree of field extension?
Let $n=[\mathbb{F}_q(\alpha):\mathbb{F}_q]$.
Since $\alpha$ is a root of $x^{q-1}-\gamma$, it follows that $n\le q-1$.
Our goal is to show that $n=q-1$.
Let $s=o(\alpha)$.
As you observed (and it's a key observation), from $o(\gamma)=q-1$ and $\alpha^{q-1}=\gamma$, we get $\alpha^{(q-1)^2}=1$, so $s{\,\mid\,}(q-1)^2$.
But in fact, $s=(q-1)^2$, as we now show . . . \begin{align*} & \alpha^s=1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \\[4pt] \implies\;& \alpha^{(q-1)s}=1 \\[4pt] \implies\;& \gamma^s=1 \\[4pt] \implies\;& (q-1){\,\mid\,}s \\[4pt] \implies\;& s=(q-1)t\;\,\text{for some positive integer $t$} \\[4pt] \implies\;& \alpha^{(q-1)t}=1 \\[4pt] \implies\;& \gamma^t=1 \\[4pt] \implies\;& (q-1){\,\mid\,}t \\[4pt] \implies\;& (q-1)^2{\,\mid\,}s \\[4pt] \implies\;& s=(q-1)^2 \\[4pt] \end{align*} as claimed.
From $[\mathbb{F}_q(\alpha):\mathbb{F}_q]=n$ we get $|\mathbb{F}_q(\alpha)|=q^n$, hence $o(\alpha){\,\mid\,}(q^n-1)$. \begin{align*} \text{Then}\;\;& o(\alpha){\,\mid\,}(q^n-1) \\[4pt] \implies\;& (q-1)^2{\,\mid\,}(q^n-1) \\[4pt] \implies\;& (q-1){\Large{\,\mid\,}}\frac{q^n-1}{q-1} \\[4pt] \implies\;& (q-1){\Large{\,\mid\,}}\sum_{k=0}^{n-1}q^k \\[4pt] \implies\;& \sum_{k=0}^{n-1}q^k\equiv 0\;\bigl(\text{mod}\;(q-1)\bigr) \\[4pt] \implies\;& \sum_{k=0}^{n-1}1^k\equiv 0\;\bigl(\text{mod}\;(q-1)\bigr) \qquad \Bigl(\text{since $q\equiv 1\;\bigl(\text{mod}\;(q-1)\bigr)$}\Bigr) \\[4pt] \implies\;& n\equiv 0\;\bigl(\text{mod}\;(q-1)\bigr) \\[4pt] \implies\;& (q-1){\,\mid\,}n \\[4pt] \implies\;& n\ge q-1 \\[4pt] \implies\;& n=q-1 \qquad \Bigl(\text{since we already have $n\le q-1$}\Bigr) \\[4pt] \end{align*} as was to be shown.