Reformulate the given system in the following form:
$$\begin{cases} x^4 + 16 = y^4 + 256 \\ x^3 - 3x^2 + 4x = 2y^3 - 12y^2 + 32y \end{cases} \quad (1) \quad (2)$$
Multiplying both sides of equation (2) with $-8$, then adding it to (1), side by side, we obtain equation:
$$(x-2)^4 = (y-4)^4 \quad (3).$$
$$(x - 2)^4 = (y - 4)^4 \quad (3).$$
$$\text{We have: } (3) \Leftrightarrow x - 2 = |y - 4| \Leftrightarrow \begin{cases} y = x + 2 \\ y = 6 - x \end{cases} \quad (4) \quad (5)$$
$$+ \text{Plugging (4) into (1), we obtain: } x^4 = (x+2)^4 + 240 \quad (6).$$
$$\text{We have: } (6) \Leftrightarrow x^3 + 3x^2 + 4x + 32 = 0 \Leftrightarrow (x+4)(x^2 - x + 8) = 0 \Leftrightarrow x = -4.$$
Consequently $y = -2$. Hence we get solution: $(x, y) = (-4, -2)$.
$$+ \text{Plugging (5) into (1), we obtain: } x^4 = (6-x)^4 + 240 \quad (7).$$
$$\text{We have: } (7) \Leftrightarrow x^3 - 9x^2 + 36x - 64 = 0 \Leftrightarrow (x-4)(x^2 - 5x + 16) = 0 \Leftrightarrow x = 4.$$
Consequently $y = 2$. Hence we get solution $(x, y) = (4, 2)$.