We have $$ x^{2006}+1 = 4y^{2006}(y+1)+2007(y+1)=(4y^{2006}+2007)(y+1). $$Now, $4y^{2006}+2007\equiv 3\pmod{4}$. Hence, there exists a prime $q\equiv 3\pmod{4}$ with $q\mid 4y^{2006}+2007$. However, for this prime, $q\mid x^{2006}+1$ is impossible, since $-1$ is not a quadratic residue modulo such a prime.