We will prove this statement by contradiction. Assume that there are positive integers $x, y, z$ satisfying the equation
$$x^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 = 0. \quad (1)$$
It is easy to check that $x \neq y$. If there are solutions of the equation (1), we have a solution of the equation (1) such that $\gcd(x, y) = (x, y) = 1$. Now let $x, y, z$ be a positive integer solution of the equation (1) such that $(x, y) = 1$.
By factoring the equation (1), we get
$$\begin{aligned} 0 &= x^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 \\ &= x^2(y^4 + 2y^2 z^2 + z^4) - y^2(x^4 - 2x^2 z^2 + z^4) \\ &= x^2(y^2 + z^2)^2 - y^2(x^2 - z^2)^2 \\ &= (x(y^2 + z^2) + y(x^2 - z^2))(x(y^2 + z^2) - y(x^2 - z^2)) \\ &= ((x - y)z^2 + x y(x + y))((x + y)z^2 + x y(y - x)). \end{aligned}$$
Then
$$(y - x)z^2 = x y(x + y) \quad \text{or} \quad (x + y)z^2 = x y(x - y).$$
By multiplying both sides of the first equation by $y - x$ we have
$$(y - x)^2 z^2 = x y(y^2 - x^2). \quad (2)$$
By multiplying both sides of the second equation by $x + y$ we have
$$(x + y)^2 z^2 = x y(x^2 - y^2) \quad (3)$$
Since we can solve the equation (3) by the same way in which we solve the equation (2), we are going to solve the equation (2).
Because $(x, y) = 1$, $(x, y^2 - x^2) = (y, y^2 - x^2) = 1$. Now the left hand side of the equation (2) is a perfect square of an integer. We know that $x, y$, and $y^2 - x^2$ are perfect squares. So there are positive integers $a, b, c$ such that $x = a^2, y = b^2, y^2 - x^2 = c^2$. Therefore we have the equation
$$b^4 - a^4 = c^2.$$
But it is well-known that there are no positive integers $a, b, c$ satisfying the above equation. It can be proved by infinite descent. Therefore, we prove the statement by contradiction. $\square$