I participated in that competition, and I had this solution but I made stupid mistake:

Bye

Here is my solution: x=0 => y=z=0. Examine the problem when $ x\geq 1$ * If x is even, $ x = 2k$. We have $ 12^x + y^4\equiv 0 (mod251)$, then $ (12^x)^{125} + (y^4)^{125}\equiv 0 (mod251)$. It means that $ A = (12^k)^{250} + (y^2)^{250}\equiv 0 (mod251)$ Use Fermat theorem, with $ (12^k,251) = 1$ and $ (y^2,251) = 1$, we have $ (12^k)^{250}\equiv (y^2)^{250}\equiv 1 (mod 251)$, so $ A\equiv 2(mod 251)$. * If x is odd. Let $ y = 2^m. t$ ( $ t$ is an odd numbers, $ m\geq 1$). At there, $ 2^{2x}.3^x + 2^{4m}.t^4 = 2^{3z}.251^z$ We can easy solve this case. Answer: $ (x,y,z) = (0,0,0)$

I have a question, if my solution is correct. Of course we first try trivial cases when one of $ {x,y,z}$ is $ 0$, which leads us to conclusion that all must be $ 0$. So, in the continuation we can assume $ x,|y|,z$ are all greater than $ 0$. We can denote $ y=2^ay_1$, where $ y_1$ is odd integer and $ a>0$. Hence the equation becomes: $ 2^{2x}3^{x}+2^{4a}y_1^4=2^{3z}251^z$ If there exists $ min\{2x,4a,3z\}$, then on dividing the equation with $ min\{2x,4a,3z\}$ we obtain an odd number on one side and even on the other. Hence $ min\{2x,4a,3z\}$ does not exist. Hence at least two numbers from $ \{2x,4a,3z\}$ are the same. Furthermore if all three were the same, we would again get one odd and one even side. Therefore exactly two are the same. Now we have three cases: 1.$ \textbf{2x=4a}$; $ \textbf{max\{2x,4a,3z\}=3z}$ $ \rightarrow a=\frac{x}{2}\rightarrow 2^{2x}(3^{x}+y_{1}^{4})=2^{3z}251^{z}\rightarrow 3^{x}+y_{1}^{4}=2^{3z-2x}251^{z}$ Now since $ 3z-2x>0, 2$ must divide $ y_{1}^{4}+3^{x}$. But since $ y_{1}^{4}\equiv1(mod 8)$ and $ 3^{2k}\equiv 1 (mod 8)$, $ 3^{2k+1}\equiv 3 (mod 8) \rightarrow y_{1}^4+3^{x}\equiv 2$ or $ 4 (mod 8)$. If $ x=2k$ we can use Fermat`s theorem as mr.danh has already shown. If $ x=2k+1 \rightarrow 4|y_{1}^4+3^{x} \rightarrow 3z-2x=2 \rightarrow z=\frac{2(x+1)}{3} \rightarrow 3|x+1 \rightarrow x=3x_{1}+1, z=2x_{1}$. Now our equation becomes: $ 2^{2}251^{2x_{1}}=y_{1}^{4}+3^{x}$, which is equivalent to $ (2\cdot251^{x_{1}}-y_{1}^{2})(2\cdot251^{x_{1}}+y_{1}^{2})=3^{x}$ Therefore $ 2\cdot251^{x_{1}}-y_{1}^{2}=3^{l}$, $ 2\cdot251^{x_{1}}+y_{1}^{2}=3^{j} \rightarrow 3^{l}+3^{j}=3s=4\cdot251^{x_{1}}$. But since $ 3$ does not divide $ 4\cdot251^{x_{1}}$ this leads us to contradiction. 2.$ \textbf{2x=3z}$; $ \textbf{max\{2x,4a,3z\}=4a}$ $ \rightarrow z=\frac{2x}{3} \rightarrow x=3x_{1} \rightarrow z=2x_{1} \rightarrow 2^{2x}(251^{2x_{1}}-3^{2x_{1}})=2^{4a-2x}y_{1}^{4} \rightarrow 251^{2x_{1}}-2^{4a-2x}y_{1}^{4}=3^{2x_{1}} \rightarrow (251^{x_{1}}-2^{2(a-x)}y_{1}^{2})(251^{x_{1}}+2^{2(a-x)}y_{1}^{2})=3^{2x_{1}}$ But this again leads us to $ 2\cdot251^{x_{1}}=3s$, which is contradiction. 3.$ \textbf{4a=3z}$; $ \textbf{max\{2x,4a,3z\}=2x}$ $ \rightarrow z=\frac{4a}{3} \rightarrow a=3a_{1} \rightarrow z=4a_{1} \rightarrow 2^{2x}3^{x}=2^{4a}(251^{z}-y_{1}^{4})$ Again $ 2^{2x-4a}3^{x}=(251^{2a_{1}}-y_{1}^{2})(251^{2a_{1}}+y_{1}^{2})$ and $ 2\cdot251^{2a_{1}}=3s$. Therefore $ (0,0,0)$ is the only solution to this equation.

tmrfea wrote: 1.$ \textbf{2x=4a}$; $ \textbf{max\{2x,4a,3z\}=3z}$ $ \rightarrow a=\frac{x}{2}\rightarrow 2^{2x}(3^{x}+y_{1}^{4})=2^{3z}251^{z}\rightarrow 3^{x}+y_{1}^{4}=2^{3z-2x}251^{z}$ Now since $ 3z-2x>0, 2$ must divide $ y_{1}^{4}+3^{x}$. But since $ y_{1}^{4}\equiv1(mod 8)$ and $ 3^{2k}\equiv 1 (mod 8)$, $ 3^{2k+1}\equiv 3 (mod 8) \rightarrow y_{1}^4+3^{x}\equiv 2$ or $ 4 (mod 8)$. If $ x=2k$ we can use Fermat`s theorem as mr.danh has already shown. If $ x=2k+1 \rightarrow 4|y_{1}^4+3^{x} \rightarrow 3z-2x=2 \rightarrow z=\frac{2(x+1)}{3} \rightarrow 3|x+1 \rightarrow x=3x_{1}+1, z=2x_{1}$. Now our equation becomes: $ 2^{2}251^{2x_{1}}=y_{1}^{4}+3^{x}$, which is equivalent to $ (2\cdot251^{x_{1}}-y_{1}^{2})(2\cdot251^{x_{1}}+y_{1}^{2})=3^{x}$ Therefore $ 2\cdot251^{x_{1}}-y_{1}^{2}=3^{l}$, $ 2\cdot251^{x_{1}}+y_{1}^{2}=3^{j} \rightarrow 3^{l}+3^{j}=3s=4\cdot251^{x_{1}}$. But since $ 3$ does not divide $ 4\cdot251^{x_{1}}$ this leads us to contradiction. Actually since $2x=4a$ then $2|x$, therefore you don't have to consider the case $2 \nmid x$.

Let $12^x+y^4=2008^z...(\star)$ it is easy to see that $y$ is even. Furthermore when $x=0$ $\vee$ $y=0$ $\vee$ $z=0$ it is easy to see that the only solution is $(x,y,z)=(0,0,0)$ If $x>0$ is even $\Longrightarrow$ $12^x+y^4\equiv 0\pmod {251}$ $\Longrightarrow$ $251|(12^\frac{x}{2})^2+(y^2)^2$, since $251$ is a prime of the form $\equiv 3\pmod 4$ we get $251|y^2$ and $251|12^{\frac{x}{2}}$ which is contradiction. If $x$ is odd, let $x=2a+1$ and $y=2^m.I$ replacing in $(\star)$ we get: $2^{4a+2}.3^{2a+1}$ $+$ $2^{4m}.I^4$ $=$ $2^{3z}.251^z$ $\Longrightarrow$ $v_2(2^{4a+2}.3^{2a+1}+2^{4m}.I^4)$ $=$ $v_2(2^{3z}.251^z)$ $\Longrightarrow$ $3z$ $=$ $4a+2$ $\vee$ $3z$ $=$ $4m$ $\Longrightarrow$ $z$ $\equiv$ $0\pmod 2$. Let $z$ $=$ $2p$ replacing in $(\star)$ we get $2^{4a+2}.3^{2a+1}$ $+$ $2^{4m}.I^4$ $=$ $2008^{2p}$ $\Longrightarrow$ $2^{4a+2}.3^{2a+1}$ $=$ $(2008^p-y^2)$ $(2008^p+y^2)$ since $2008^p+y^2$ $\equiv$ $1,2\pmod 3$ we get $3^{2a+1}|2008^p-y^2$ $\Longrightarrow$ $(2008^p+y^2)|2^{4a+2}$ $\Longrightarrow$ $(2008^p+y^2)$ $=$ $2^\alpha$ with $\alpha\leq 4a+2$ $\Longrightarrow$ $2^{3p}.251^p$ $+$ $2^{2m}.I^2$ $=$ $2^\alpha$, if $3p$ $\neq$ $2m$ $\Longrightarrow$ $\alpha$ $=$ $\min\{2m,3p\}$ but $2^{3p}.251^p$ $+$ $2^{2m}.I^2$ $>$ $2^{\min\{2m,3p\}}$ $=$ $2^\alpha$ $\Longrightarrow$ $3p$ $=$ $2m$ $=$ $r$ $\Longrightarrow$ $p$ is even, let $p=2p'$ $\Longrightarrow$ $2^r(251^{2p'}+I^2)$ $=$ $2^\alpha$ $\Longrightarrow$ $2^{\alpha-r}$ $=$ $251^{2p'}+I^2\equiv 2\pmod 4$ $\Longrightarrow$ $\alpha-r=1$ $\Longrightarrow$ $251^{2p'}+I^2=2$ which is contradiction. Hence all solutions are $(x,y,z)=(0,0,0)$.