The Wilcoxon-Mann-Whitney makes no distributional assumptions and so could always be applied. It has the advantage of providing exceptionally good robustness; for instance, if there were to be a few extremely bad points in your data (e.g., individuals with 10000+ diseases) these wouldn't heavily influence the Wilcoxon-Mann-Whitney test, whereas even one of these would destroy the t-test. Having said that, the t-test is also quite robust, especially for large sample sizes, just not to the same extent as the Wilcoxon-Mann-Whitney test. For large sample sizes, such as in your scenario, the t-test does not require an assumption that the data be normally distributed; it only needs to come from a distribution to which the Central Limit Theorem applies (i.e., one with finite variance). The main advantage of the t-test is that it typically provides somewhat higher power.
The two tests also address somewhat different hypotheses. For the t-test, the null hypothesis is that the two populations have equal means ($H_0: \mu_1 = \mu_2$), whereas for the Wilcoxon-Mann-Whitney test, the null hypothesis is that given random representatives $X$ and $Y$ from the first and second populations respectively, there is an equal probability that $X$ is greater than $Y$ compared to the other way around ($H_0: P(X>Y) = P(Y>X)$).
In the end, there are justifications for going with either test. It might come down to what is expected for your field/audience. Personally, I would probably go with the t-test, on account of its slightly greater power and more straightforward interpretation, assuming that an inspection of the data (say, looking at the two histograms) does not reveal any red flags (i.e., heavy tails or implausibly large values).