If $\alpha$ is not algebraic over $Z_{p}$ then $x^p - \alpha$ is inseparable over $Z_{p}(\alpha)$ to prove this I supposed $x^p - \alpha$ is separable. Since $char(Z_{p}(\alpha))=p$ and by the form of the polynomial we can conclude through a theorem that $x^p - \alpha$ can't be irreducible, so it is reducible. Any hints to complete?
Note that I'm working here with the old definition of separability.
The point here is that we are working with the "old definition of separability". We have the following theorem:
Suppose that $f\in F[x]$ is irreducible over $F$. Then $f$ is inseparable over $F$ if and only if $char(F)$ is a prime, say p, and $f(x)$ has the following form: $f(x)=a_{0} +a_{1}x^p +...+a_{n}x^{n.p}$.
In our care, if $x^p - \alpha$ is irreducible then by applying the above theorem for $F=Z_{p}(\alpha)$ we can get directly that $x^p - \alpha$ is inseparable.
If not, then we can write: $x^p - \alpha = g(x)h(x)$ where $g$ is irreducible over $Z_{p}(\alpha)$. Let $\beta$ to be a root of $g$ in $K$:=the splitting field of $x^p - \alpha$, then $\beta$ is also a root for $x^p - \alpha$, thus $\beta^p=\alpha$. We have now $x^p-\alpha = x^p -\beta^p=(x-\beta)^p$. So we conclude that $g(x)=(x-\beta)^i$ for some $1\leq i \leq p-1$. Since $g$ irreducible, then if $2\leq i$ then $g$ is clearly inseparable. If not then $g(x)= x-\beta$. Thus $\beta \in Z_{p}(\alpha)$, i.e. $\beta=m(\alpha)/n(\alpha)$ for some polynomials $m,n \in Z_{p}[x]$ and such that $n(\alpha)$ is not zero. Take now $t(x):=n(x^p).x-m(x^p)$. It is clear that $t\in Z_{p}[x]$ and $t(\beta)=0$. Therefore $\beta$ is algebraic over $Z_{p}$ and so is $\alpha$, contradiction. So this case is not possible and the proof is completed.