After manipulation equation becomes
$$\begin{align*} x^2 y^2 (x-y)^2 - 6^2 &= 0, \quad x, y \in \mathbb{Z} \\ &\Leftrightarrow [xy(x-y)-6][xy(x-y)+6] = 0 \\ &\Leftrightarrow xy(x-y) = 6, \quad x, y \in \mathbb{Z} \text{ or } xy(x-y) = -6, \quad x, y \in \mathbb{Z} \\ &\Leftrightarrow xy(x-y) = 6, \quad x, y \in \mathbb{Z} \text{ (1) or } xy(y-x) = 6, \quad x, y \in \mathbb{Z} \text{ (2)} \end{align*}$$
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From (1) and (2) it is clear that, if $(x_0, y_0)$ is a solution of (1), then the pair $(y_0, x_0)$ is a solution of (2) and vice versa. Thus is enough to solve equation (1).
Since $x, y \in \mathbb{Z}$, equation (1) is equivalent to:
$$\begin{array}{l} \{xy = 6, x - y = 1\} (\Sigma_1) \text{ or } \{xy = -6, x - y = -1\} (\Sigma_2) \\ \text{or } \{xy = 3, x - y = 2\} (\Sigma_3) \text{ or } \{xy = -3, x - y = -2\} (\Sigma_4) \\ \text{or } \{xy = 1, x - y = 6\} (\Sigma_5) \text{ or } \{xy = -1, x - y = -6\} (\Sigma_6) \\ \text{or } \{xy = 2, x - y = 3\} (\Sigma_7) \text{ or } \{xy = -2, x - y = -3\} (\Sigma_8). \end{array}$$
From the eight systems only $(\Sigma_1), (\Sigma_3), (\Sigma_8)$ have integer solutions:
$$\begin{aligned} (x, y) &= (3, 2), \quad (x, y) = (-2, -3), \quad (x, y) = (3, 1), \\ (x, y) &= (-1, -3), \quad (x, y) = (-2, 1), \quad (x, y) = (-1, 2). \end{aligned}$$
According to the previous discussion equation (2) has also the solutions:
$$\begin{aligned} (x, y) &= (2, 3), \quad (x, y) = (-3, -2), \quad (x, y) = (1, 3), \\ (x, y) &= (-3, -1), \quad (x, y) = (1, -2), \quad (x, y) = (2, -1). \end{aligned}$$