What is the maximum number of intersection points of two different Bernoulli lemniscates in the real plane? (Of course two identical lemniscates share an infinite number of points.)

Here are some of my efforts: a Bernoulli lemniscate is a degree four curve with two nodes on the line of infinity in complex projective plane: $I=(i:1:0), J=(-i:1:0)$. By Bezout's theorem every two of these curves intersect in 16 points (counted with multiplicity) in $\mathbb{P^2(C)}$. But two common nodes in $I,J$ contribute at least (and exactly in the generic case) $8 = 2\times 2 + 2\times 2$ in the intersections. So the number of real intersections (i.e. intersections in $\mathbb R^2$) is at most $8$.
However, according to my experience with Mathematica, I think that the maximum number of intersections is at most 6. So what is the correct answer to this problem? If the answer is 6, where are the other two complex points, and can we prove their existence algebraically? (Note I'm interested in the algebraic approach to this problem and not analytic or topological solutions.)
Thanks!