\[x^{y}+y^{x}=z^{y}\:\: ...(1)\] \[x^{y}+2012=y^{z+1}\:\: ...(2)\] We can easily check that x, y, z>1. Case 1. x is even By (2), y is even. And by (1), z is even. z+1≥3 so 8 divides y^(z+1) and x^y≡4 (mod 8). ∴y=2, x≡2 (mod 4) x^2+2^x=z^2 <=> 2^x=(z+x)(z-x). Let z+x=2^(a+1), z-x=2^(b+1). Then x=2^a-2^b≡2 (mod 4) so b=1. 2^a-2=x=(a+1)+(b+1)=a+3 so a=3, x=6, z=10. (x, y, z)=(6, 2, 10) is a solution. ∴(x, y, z)=(6, 2, 10) Case 2. x is odd By (2), y is odd. And by (1), z is even. By (1), x+y≡0 (mod 4). By (2), x≡y (mod 4). So we get x, y is even, which is contradiction. By Case 1 and 2, the only solution is (x, y, z)=(6, 2, 10).

That $2012$ is obviously there to destroy large $v_2$s appearing. So let's start with $x$ even. If $x$ is even and $y \ge 3$, the L.H.S is $4$ in mod 8, so $z+1 \le 2$, so $z=1$, and the first equality fails. So $y \le 2$, and obviously $y>1$, so $y=2$. Now we have $x^2 + 2^x = z^2$ and $x^2 + 2012 = 2^{z+1}$. $x=6$ and $y=10$ works, but since $x^2 + 2^x = z^2= ( \log_2 (x^2+2012) - 1)^2$, obviously larger $x$ fails by bounding. If $x$ is odd, $y$ is odd as well, so $z$ is even. Since $y > 1$, $z^y$ is a multiple of $4$, so $x+y$ is a multiple of $4$. Then working mod 4 on the second equality gives us the desired contradiction.

It is easy for $x$ and $y$ to be evan. Now we see the odd state of x and y.hence z is a evan. $$ x^y + y^x \equiv x+y \equiv 0 (mod4)$$$$ x^y + 2012 \equiv x \equiv y (mod4)$$contradiction