Taiwan IMO Selection Camp

First note that $x$ divides $2012 \cdot 2 = 2^3 \cdot 503$. If $503 \mid x$ then the right-hand side of the equation is divisible by $503^3$, hence $503^2 \mid xyz + 2$. Since $503 \mid x$, this is a contradiction. Therefore $x = 2^m, m \in \{0, 1, 2, 3\}$. If $m \ge 2$ then $2^6 \mid 2012(xyz + 2)$. However, the highest power of $2$ dividing $2012$ is $2^2$, and the highest power of $2$ dividing $xyz + 2 = 2^m yz + 2$ is $2^1$. So $x = 1$ or $x = 2$ produces the following two equations
$$y^3 + z^3 = 2012(yz + 2) \quad \text{and} \quad y^3 + z^3 = 503(yz + 1).$$
In both of the equations above, we obtain that the prime $503 = 3 \cdot 167 + 2$ divides $y^3 + z^3$. We claim that $503 \mid y+z$. If $503 \mid y$, then $503 \mid y+z$ is obvious. So we assume that $503 \nmid y$ and $503 \nmid z$. Then by Fermat's Little Theorem we get $y^{502} \equiv z^{502} \pmod{503}$. On the other hand, $y^3 \equiv -z^3 \pmod{503}$ implies $y^{3 \cdot 167} \equiv -z^{3 \cdot 167} \pmod{503}$, that is, $y^{501} \equiv -z^{501} \pmod{503}$. Hence we obtain $y \equiv -z \pmod{503}$; that is, $503 \mid y+z$.
Therefore $y+z=503k, k \ge 1$. In view of $y^3+z^3 = (y+z)((y-z)^2+yz)$, the two equations above can respectively be put into the following forms
$$k(y - z)^2 + (k - 4)yz = 8, \qquad (1)$$
$$k(y - z)^2 + (k - 1)yz = 1. \qquad (2)$$
In (1), we have $(k-4)yz \le 8$, from which we get $k \le 4$. Indeed if $k > 4$, then $1 \le (k-4)yz \le 8$, which forces $y \le 8$ and $z \le 8$. This is impossible, because $y+z = 503k \ge 503$. Next note that in the first equation $y^3+z^3$ is even. Therefore $y+z = 503k$ is also even, meaning that $k$ is even. So $k=2$ or $k=4$. Clearly when $k=4$, equation (1) has no integer solutions. If $k=2$ then equation (1) can be written as $(y+z)^2 - 5yz = 4$. Since $y+z = 503k = 503 \cdot 2$, we get $5yz = 503^2 \cdot 2^2 - 4$. However $503^2 \cdot 2^2 - 4$ is not a multiple of $5$. Therefore equation (1) has no integer solutions.
In (2), we have $0 \le (k-1)yz \le 1$, from which we get $k=1$ or $k=2$. Also $0 \le k(y-z)^2 \le 1$, so $k=2$ only when $y=z$. In that case $y=z=1$, which is false since $y+z \ge 503$. Therefore $k=1$ and equation (2) can be written as $(y-z)^2 = 1$, giving $z-y = |y-z| = 1$. From $k=1$ and $y+z = 503k$, we deduce $y = 251, z = 252$. In summary, the triple (2, 251, 252) is the unique solution.