Let $F$ be a field with $|F|=q$ and $[K:F]=2$. Let $α∈F$ of order $q-1$. Then there exist an element $β \in K$ of order $q^2-1$ such that $β^{q+1}=α$

Question

Let $F$ be a field with $|F|=q$ and $[K:F]=2$. Let $α∈F$ of order $q-1$. Then there exist an element $β \in K$ of order $q^2-1$ such that $β^{q+1}=α$

Am stuck with finding such a $\beta$ of order $q^2-1$. However since $\gcd(q+1,q^2-1)=q+1$ and $α^{(q^2-1)/(q+1)}=α^{q-1}=1$, I can assure that $x^{q+1}=α$ has solution in $K$. But how to find such an element of order $q^2-1$?

Please help me to proceed! Thanks in advance.