Self-intersecting walk in expander graphs
I think only knowing the graph is an expander does not fix the value you are asking. If you are doing random walk naively then when the graph has bounded degree there is constant probability that the walk will reach the same vertex after just 2 steps, and therefore the expected number of steps is constant.

I guess you want to exclude that case by saying the random walk cannot take the same edge in two consecutive steps. But still there are different cases here: if we take the union of a degree 4 expander graph and a grid, it will still be an expander (with worse parameters), but there is constant probability that the walk return to the same position in 4 steps, and therefore the expected number of steps is again constant. On the other hand, if we take a random graph, since the girth is high the expected number (or in fact the minimum) of steps for any vertex to be revisited will be super-constant.

$\sqrt{n}\log n$ is definitely a safe upperbound because of the birthday paradox argument. I guess the answer for random graphs might be $\sqrt{n}$, because if you take a random walk of length significantly smaller than $\sqrt{n}$, with good probability the induced subgraph will be just the path and do not have more edges. In general if the induced graph on the set of visited vertices has a lot of edges outside the path, it means the walk "had many chances" to visit some vertex twice.

I understand this may not answer your question completely but I couldn't figure out how to leave a comment like Suresh Venkat did.