Uncorrelated Bernoulli and Poisson variables need not be independent ($\mathrm{Cov}(X,Y)=p(e^{-p}-p),p=e^{-p}⟹\mathrm{Cov}(X,Y)=0$)

Here is an explicit joint distribution, for $0<p<0.8$: assign mass $(1+p)e^{-p}-p$ at $(0,0)$, mass $p(1-e^{-p})$ at $(1,0)$, mass $p{e}^{-p}$ at $(1,1)$, and mass $e^{-p}{p}^k/k!$ at $(0,k)$ for $k\ge 2$; all other masses are zero. These are nonnegative: the first is decreasing in $p$, and at $p=0.8$ it is still positive. Summing over rows and columns gives a Bernoulli distribution with mean $p$ and a Poisson distribution with mean $p$, so the joint mass sums to one.

Only $(1,1)$ contributes to the mixed first moment, giving $\mathbb{E}[XY]=p{e}^{-p}$ and the covariance above. The strictly decreasing function $e^{-p}-p$ has a unique zero in $(0,0.8)$, approximately $0.567143$. At that parameter the variables are uncorrelated. They are nevertheless dependent, because $\mathbb{P}(X=1\mid Y=1)=1$ whereas $\mathbb{P}(X=1)=p<1$. The Bernoulli independence from zero covariance result requires both variables to be Bernoulli and does not apply here.