$ x=0$ or $ y=0$ are not solutions $ y^x(14^x+y^y)=2007=3^2\cdot 223 \Rightarrow y\in \{ 1,3,3^2,223,223\cdot 3,223 \cdot 3^2 \}$ Finally, the only possible solution is $ (x,y)=(2,3)$

$ y = 0,x = 0$ are not solutions and $ 14^3 = 2744$ so $ x = \{1,2\}$ and $ x = 1$ not solution so $ x = 2$ and $ y = 3$

dima ukraine wrote: $ y = 0,x = 0$ are not solutions and $ 14^3 = 2744$ so $ x = \{1,2\}$ and $ x = 1$ not solution so $ x = 2$ and $ y = 3$ I solved it the same way using that $ y$ is odd and $ y^{y + 1}\le2007 , y^{y + 2}\le2007$ for $ y = 1,3$