Solution:

If $p = 3$, then $p^{2} + 2007p - 1 = 6029$ which is a prime.

For $p \neq 3$, we know that $p = 3k \pm 1$ for some integer $k$. Hence
$$p^{2} + 2007p - 1 = (3k \pm 1)^{2} + 2007(3k \pm 1) - 1 = 9k^{2} \pm 6k + 1 + 2007 \cdot 3k \pm 2007 - 1 = 9k^{2} \pm 6k + 2007 \cdot 3k \pm 2007.$$
This expression is divisible by $3$ and can't be prime (unless it equals $3$, but for $p > 3$ it is much larger than $3$). Thus the only such prime number is $p = 3$.