The given equality is equivalent to
$$\begin{aligned} 6x^2y^2 + 3x^2 - 4y^2 &= 2012 \\ 3x^2(2y^2 + 1) - 2(2y^2 + 1) &= 2012 - 2 \\ (3x^2 - 2)(2y^2 + 1) &= 2010 \end{aligned}$$
Since $2010 = 2 \cdot 3 \cdot 5 \cdot 67$ and the number $2y^2 + 1$ is odd, we conclude that
$$\begin{aligned} 2y^2 + 1 &\in \{1, 3, 5, 15, 67, 201, 335, 1005\}, \\ 2y^2 &\in \{0, 2, 4, 14, 66, 200, 334, 1004\}, \\ y^2 &\in \{0, 1, 2, 7, 33, 100, 167, 502\}. \end{aligned}$$
Thereby $y^2$ must be $0$, $1$ or $100$.
There are three cases left:
$$\left\{ \begin{array}{l} 3x^2 - 2 = 2010 \\ 2y^2 + 1 = 1 \end{array} \right. \quad \left\{ \begin{array}{l} 3x^2 - 2 = 670 \\ 2y^2 + 1 = 3 \end{array} \right. \quad \left\{ \begin{array}{l} 3x^2 - 2 = 10 \\ 2y^2 + 1 = 201 \end{array} \right.$$
We see that only in the third case there are integer solutions,
$$(x, y) \in \{(2, 10), (-2, 10), (2, -10), (-2, -10)\}.$$