There exist some general results which show when a product of Baire spaces is Baire. Specifically, Theorem 2 here states

If $X,Y$ are two Baire spaces and at least one has a *locally countable pseudo-base*, then $X\times Y$ is a Baire space.

A *pseudo-base* of a topological space $X$ is a family of nonempty open sets such that every open set in $X$ contains an element of the family. In particular, any base is a pseudo-base. A family is *locally countable* if any element of the family contains only countably many other ones. In particular, any countable family is locally countable.

Now, any Bernstein set, in fact any subset of a second countable space, is second-countable, which means it has a countable base, which is a locally countable pseudo-base. Therefore, by the theorem, a product of Bernstein sets is Baire.

It is worth noting that not every product of Baire spaces is Baire. Oxtoby (the author of the linked paper) constructs a Baire space whose square is not Baire assuming the continuum hypothesis. Cohen has shown CH is not necessary for that