Let $\gcd(x, y) = d$ and $x = ad$, $y = bd$. Then the equation can be written as $\frac{a^2+249ab+b^2}{a^2b^2d^2} = \frac{1}{2012}$ or
$$a^2 b^2 d^2 = 2012(a^2 + 249ab + b^2).$$
As $a$ and $b$ are relatively prime, $a^2$ and $b^2$ are both relatively prime to $a^2 + 249ab + b^2$ and therefore both they must be divisors of $2012$. As $2012 = 2^2 \cdot 503$ and $503$ is a prime, the possible cases are $(a, b) = (1, 1)$, $(a, b) = (1, 2)$, $(a, b) = (2, 1)$. If we substitute $(a, b) = (1, 1)$ into the last equation, we get $d^2 = 2012 \cdot 251$, which is not solvable in integers. The other two cases give $4d^2 = 2012 \cdot 503$, from which $d = 503$. This leads to the solutions $(x, y) = (503, 1006)$ and $(x, y) = (1006, 503)$.

$$2012x^2 + 249 \cdot 2012xy + 2012y^2 = x^2y^2.$$
From the left-hand side we see that both sides of the equation must be divisible by $503$. As $503$ is a prime, one of the numbers $x$ and $y$ must be divisible by $503$. So $x^2$ or $y^2$ is divisible by $503^2$, giving that both sides of the equation are divisible by $503^2$. If $x$ is divisible by $503$, the summands $2012x^2$ and $249 \cdot 2012xy$ on the left-hand side are divisible by $503^2$, meaning that $2012y^2$ is divisible by $503^2$. Therefore $y$ is divisible by $503$. Analogously, we get that if $y$ is divisible by $503$, then $x$ is also divisible by $503$. Consequently, both $x$ and $y$ are divisible by $503$. Denote $x = 503a$, $y = 503b$. Then the equation, after dividing both sides by $503^3$, simplifies to $4a^2 + 996ab + 4b^2 = 503a^2b^2$. Assume $a \ge b$. If $b \ge 2$, then $503a^2b^2 \ge 503a^2 \cdot 2b = 1006a^2b = 4a^2b + 4a^2b + 998a^2b > 4a^2 + 4b^2 + 996ab$, so the last equation cannot hold. Therefore $b = 1$. Now we get a quadratic equation $499a^2 - 996a - 4 = 0$ with respect to $a$, whose only positive solution is $a = 2$. From here we obtain the solution $(1006, 503)$ to our original equation. The case $b \ge a$ is symmetrical and gives the solution $(503, 1006)$.