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 $4 y^{2006}+2007 \equiv 3(\bmod 4)$, so $x^{2006}+1$ will have at least one prime divisor of the type $4 k+3$. It is known (and easily obtainable by using Fermat's Little Theorem) that this is impossible.

1. Start with the given equation:
$$x^{2006} - 4y^{2006} - 2006 = 4y^{2007} + 2007y$$

2. Rearrange the equation to isolate $x^{2006}$:
$$x^{2006} = 4y^{2006} + 4y^{2007} + 2007y + 2006$$

3. Factor the right-hand side:
$$x^{2006} + 1 = 4y^{2006}(y + 1) + 2007(y + 1)$$
$$x^{2006} + 1 = (4y^{2006} + 2007)(y + 1)$$

4. Consider the expression $4y^{2006} + 2007$. Notice that:
$$4y^{2006} + 2007 \equiv 3 \pmod{4}$$
This is because $4y^{2006} \equiv 0 \pmod{4}$ and $2007 \equiv 3 \pmod{4}$.

5. Since $4y^{2006} + 2007 \equiv 3 \pmod{4}$, there exists a prime $q \equiv 3 \pmod{4}$ such that $q \mid 4y^{2006} + 2007$.

6. Now, consider the prime $q$ and the expression $x^{2006} + 1$. If $q \mid 4y^{2006} + 2007$, then $q \mid x^{2006} + 1$.

7. However, for a prime $q \equiv 3 \pmod{4}$, $-1$ is not a quadratic residue modulo $q$. This means that there is no integer $a$ such that $a^2 \equiv -1 \pmod{q}$.

8. Therefore, $x^{2006} + 1 \equiv 0 \pmod{q}$ implies that $x^{2006} \equiv -1 \pmod{q}$, which is impossible because $-1$ is not a quadratic residue modulo $q$.

9. Hence, there is no solution in the set of positive integers for the given equation.

$\blacksquare$