The system does not have a solution since adding the equation gives $-(x - y)^4 = 4042$ whose l.h.s. is non-positive but r.h.s. is positive.
We show that the first equation $4x^3y - x^4 - 3x^2y^2 = 2021$ has solutions (the same could be done for the second equation by symmetry). Dividing the equation by $x^4$ and reordering the terms in the l.h.s. results in
$$-3 \left(\frac{y}{x}\right)^2 + 4 \left(\frac{y}{x}\right) - 1 = \frac{2021}{x^4}. \qquad (6)$$
As the discriminant of the quadratic equation $-3t^2 + 4t - 1 = 0$ is positive, there exists a real number $t$ such that $-3t^2 + 4t - 1$ equals a positive number $\varepsilon$. Define $x = \sqrt[4]{\frac{2021}{\varepsilon}}$ and $y = xt$; then $x$ and $y$ satisfy (6) and also the first equation of the given system of equations. Hence Anne is right.