Consider the function $f(x,y) = xy$ for $x,y>0$. Isn't $f$ a convex function? I computed the Hessian to be a matrix with only off diagonal entries equal to one and others zero.
For any vector $z$ belonging to the domain $z^T H z$ is clearly positive. Doesn't this suggest $f$ is convex??
No, it's not convex, and no, $z^T H z$ is not positive. Try $z = \pmatrix{1\cr -1}$. Note: in the criterion for convexity involving the Hessian, there's no requirement for the vector $z$ to belong to the domain of $f$. The vector $z$ does not correspond to a point in the domain, but rather to an infinitesimal displacement that can be in any direction.
EDIT: To see directly that $f$ is not convex, note that $f(1,3) = f(3,1) = 3$, $(2,2) = \frac{1}{2} (1,3) + \frac{1}{2} (3,1)$, but $f(2,2) = 4 > 3$.