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