Proposition 5.21 in Atiyah-MacDonald

If your doubt is what I think it is, then remember: if $\,k/F\,$ is a fields extension and $\,w\in k\,$, then $\,F(w)=F[w]\Longleftrightarrow w\,$ is algebraic over $\,F\,$.

In you case, since $\,k\,$ is a field, it then equals the polynomial ring in $\,k[\overline x]\,$ iff this element is algebraic over $\,k\,$

Suppose $\bar x$ is not algebraic over $k$. Then $k[\bar x]$ is isomorphic to the polynomial ring $k[X]$. However $k[X]$ is not a field(any polynomial of degree $\geq 1$ in $k[X]$ is not invertible in $k[X]$). Hence $k[\bar x]$ is not a field. This is a contradiction. Hence $\bar x$ is algebraic over $k$.