This answer is essentially the same as the comment above. Let $d$ be any positive integer that is prime to the characteristic of $k$. Let $R$ be the $k$-algebra, $$ R=k[x_0,x_1,x_2,\dots,x_n,\dots]/\langle x_0x_1 - 1, x_2^d - x_1, x_3^d - x_2,\dots, x_{n+1}^d-x_n,\dots \rangle. $$ For every integer $n$, define $R_n\subset R$ to be the $k$-subalgebra generated by $x_0,x_1,\dots,x_n$. Then $R_n$ is isomorphic to $k[x_n,x_n^{-1}]$, which is a smooth $k$-algebra. Moreover, the transition map $R_n\to R_{n+1}$ is the same as $$ f_n : k[x_n,x_n^{-1}] \to k[x_{n+1},x_{n+1}^{-1}], \ \ f(x_n) = x_{n+1}^d. $$ This is étale since $d$ is prime to the characteristic. The ring $R$ is not Noetherian