Finding ("Nonadaptive" means you have to specify all the tests in advance, whereas "adaptive" means you can use the results from previous tests before deciding which ones to do next.)
The standard reference on group testing appears to be Combinatorial Group Testing and Its Applications, by Du and Hwang. Part II, which comprises Chapters 7-9, is on nonadaptive testing. In particular, finding optimal testing structures when there are two or more "defectives" is still an open problem.
However, if $t(d,n)$ is the number of tests required to isolate $d$ defectives out of $n$ total subjects, the bounds $\Omega(\frac{d^2}{\log d} \log n) \leq t(d,n) \leq O(d^2 \log n)$ are known. The Wikipedia article on disjunct matrices has a discussion and some proofs.
It might be interesting to compare the solution for the adaptive version of this problem, as we can give a definite answer in this case.
Let $n(t)$ denote the maximum number of bottles of wine for which 2 poisoned ones can be identified in $t$ adaptive tests. In "Group testing with two and three defectives" (Annals of the New York Academy of Sciences 576, pp. 86-96, 1989) Chang, Hwang, and Weng give explicit testing procedures that yield the lower bounds $$n(t) \geq 89 \cdot 2^{\frac{t}{2}-6}, t \text{ even, } t \geq 12;$$$$n(t) \geq 63 \cdot 2^{\frac{t-1}{2}-5}, t \text{ odd, } t \geq 13.$$
In the Du and Hwang text it is shown that, for $t \geq 4$, we have the upper bound $$n(t) \leq 2^{\frac{t+1}{2}} - 1/2.$$
(Note that this is the upper bound on $f(n)$ given in Sergey Norin's answer.)
These bounds tell us that $n(18) \leq 723$ but that $n(19) \geq 1008$. Thus 2 poisoned bottles can be identified out of 1000 in 19 adaptive tests but no fewer, using the testing procedure described in the Chang, Hwang, and Weng paper.
