Theorem: Let $K$ be an infinite field and let $L:=K(\alpha, \beta)/K$ be a field extension with $\alpha$ algebraic over $K$ and $\beta$ separable over $K$. Then $L = K(z)$ for a certain $z \in L$.
Proof: Let $f,g$ be the minimal polynomials over $K$ of $\alpha, \beta$ respectively and $M/L$ the splitting field of $fg$.
Denote by $\alpha_1, \ldots, \alpha_m$ the roots of $f$ in $M$ and $\beta_1, \ldots, \beta_n$ the roots of $g$ in $M$.
Since $K$ is infinite, there exists a $c \in K$ such that $c\alpha_i+\beta_j$ are all different. Pose $z := c\alpha+\beta \in L$ and $h(x) := g(z-cx) \in K(z)[x]$.
Then $h(\alpha) = 0 = f(\alpha)$ and $(x-\alpha)|\gcd(f,h)$. It can be verified that $\alpha$ is the only common root of $f$ and $h$. Since $f$ is separable, $\gcd(f,h) = x-\alpha$.
Now $x-\alpha \in K(z)[x]$ so in particular $\alpha \in K(z)$ and $\beta = z-c\alpha \in K(z)$.
Therefore... $\square$
Question: A priori, we may seek $\gcd(f,h)$ either in $M[x]$ or in $K(z)[x]$. But $\alpha$ is not known to be in $K(z)$ yet, so what tells us that the $\gcd$ of $f$ and $h$ in $K(z)[x]$ is also $x-\alpha$ and not, say, $1$?
Just look at $f$ and $h$ in some splitting field. They have $x-\alpha$ as a common factor. Now the gcd can be calculated algebraically from $f$ and $g$ by Euclid's algorithm. Thus this gcd will be in any field that contains the coefficients of $f$ and $g$.