Let $k$ be an algebraically closed field. Glue two copies of $\text{Spec}(k[[x]])$ along $\text{Spec}(k((x)))$. This gives a scheme $X = U \cup V$ such that any etale covering of $X$ can be refined by the Zariski covering $X = U \cup V$. (Hint: use that $k[[t]]$ is strictly henselian.) Thus the etale Cech cohomology is the usual Cech cohomology. For a constant sheaf with value an abelian group $A$ we get $\check{H}^i_{et}(X, A) = 0$ for $i > 0$. On the other hand, if $A$ is finite of order invertible in $A$ then we have $H^2_{et}(X, A) \cong A$ as you can see by using the Cech to cohomology spectral sequence and using the fact that $H^1_{et}(U \cap V, A) \cong A$ as abelian groups (noncanonically).

Currently I am blanking on a separated example. If I remember, I will post another answer.

Let $k$ be an algebraically closed field. Glue two copies of $\text{Spec}(k[[x]])$ along $\text{Spec}(k((x)))$. This gives a scheme $X = U \cup V$ such that any etale covering of $X$ can be refined by the Zariski covering $X = U \cup V$. (Hint: use that $k[[t]]$ is strictly henselian.) Thus the etale Cech cohomology is the usual Cech cohomology. For a constant sheaf with value an abelian group $A$ we get $\check{H}^i_{et}(X, A) = 0$ for $i > 0$. On the other hand, if $A$ is finite of order invertible in $A$ then we have $H^2_{et}(X, A) \cong A$ as you can see by using the Cech to cohomology spectral sequence and using the fact that $H^1_{et}(U \cap V, A) \cong A$ as abelian groups (noncanonically).

Currently I am blanking on a separated example. If I remember, I will post another answer.

Edit: In the paper by Stefan Schoer, "Geometry on totally separably closed schemes", the author conjectures that for separated and quasi-compact schemes etale Cech cohomology should agree with etale cohomology. (See remarks at the top of page 540.) He even conjectures it should suffice if the diagonal is affine. I have no idea how reasonable this conjecture is.