Suppose I observe 0 successes out of $n$ binomial trials.
A Wald test would have me compute
$$ \dfrac{(\hat{\pi} - \pi_0)^2}{\widehat{V}(\hat{\pi}_j)} \sim X^2_1$$
But the estimated variance is 0. Is a Wald test not appropriate for cases where no successes (or failures) are observed?
I cannot think of a single reason, other than perhaps ignorance, to use the Wald test or Wald interval for this problem. The fact that, as you noted, it can lead to a zero estimated variance is one of the problems with it.
The default test is the score test; the Wald method is sometimes used to build intervals. One way to repair the Wald interval in this case is to add two successes and two failures. This approximates, but is slightly wider than, the score interval, and generally has good coverage properties.
