Let $X$ be a proper algebraic variety. $X$ is said to have polynomial point count if there is a polynomial $P$ such that for all finite fields $\mathbb F_q$ with $q$ elements, $|X(𝔽_q)|=P(q)$.

If in addition $X$ is smooth, then by the Weil conjectures one can derive that $X$ has no odd cohomology.

My question: Is there a weaker hypothesis than smooth under which the same conclusion can be reached?

Please don't answer with the condition $X$ has a paving by affine cells. Because I know about this condition, and know about it I should, since many (but not all) varieties that one comes across in real life (by which I provocatively and rather tounge-in-cheek mean my own mathematical experience) have affine pavings.

It wouldn't surprise me if there is no answer to this question as stated. So I would also like to issue a call for examples. In particular, what is the best-behaved variety you know which has polynomial point count and odd cohomology?

The example I know is the nodal plane cubic $$Y^2Z=X^2(X+Z)$$. It has $q$ points over $𝔽_q$ and one-dimensional $H^1$. Can you do better?

(If you are at all concerned that I haven't defined what base scheme $X$ is over or what cohomology theory I'm using, feel free to interpret such things in any sensible way you wish)