, 2007

Solution:
We assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have
$$\begin{gathered} x^{2006} = 4 y^{2007} + 4 y^{2006} + 2007 y + 2006 \\ x^{2006} + 1 = 4 y^{2006}(y + 1) + 2007(y + 1) \\ x^{2006} + 1 = \left(4 y^{2006} + 2007\right)(y + 1) \end{gathered}$$
But $4 y^{2006} + 2007 \equiv 3 \pmod{4}$, so $x^{2006} + 1$ will have at least one prime divisor of the type $4k + 3$. It is known (and easily obtainable by using Fermat's Little Theorem) that this is impossible.