First, $a_2^2 - a_1^2 = a_{1}a_{3}$ implies $a_3=-1$. Easy to prove that $a_n = \pm 1$ for odd $n$ and $a_{n+2} = -a_n$ for odd $n$; it all follows by induction from $a_n ^2 - a_2^2 = a_{n-2}a_{n+2}$. Therefore $(a_1, a_3, a_5, a_7, \dots) = (1, -1, 1, -1, \dots)$ thus $a_{2007} = -1$. And to prove the existence of such sequence, set $a_n = 0$ for even $n$; then it's easy to prove that $a_n^2 - a_m^2 = a_{n-m}a_{n+m}$ for all $n, m$ by looking at the cases $n, m$ odd, even.