Plane quartic with exactly two nodes ($(X^2-Z^2{)}^2+XZ{Y}^2+Y^4=0$)

Over the complex numbers the displayed projective plane curve has precisely two singular points, $[1:0:1]$ and $[-1:0:1]$. At either, with $X=\pm 1+u$, $Y=v$, $Z=1$, the leading quadratic term is $4u^2\pm v^2$, so they are ordinary double points. The derivative equations in the chart $Y=1$ have no solution: writing $d=X^2-Z^2$ forces both $d^2=-1/16$ and $d^2=1$. The Bézout theorem proves irreducibility, since a factorization into degrees $r,4-r$ would require at least three intersections, whereas the only possible component intersections are the two transverse ordinary double points.