Takes just 10 minutes to prove it!

Let's assume that $K=2007+4^n$ . Thus K is an odd number and it can be expressed in the form: $2k+1$ . Suppose for any positive integer $n$ we would get some $k$ such that,

$2007+4^n=(2k+1)^2$ which yields that, $2006=4(k(k+1)-4^{n-1})$ So it says $4$ divides $2006$ , which isn't possible, a contradiction.

The expression when is divided by 4 has a remainder of 3, for all natural number n. In the other hand, we know that a perfect square when is divided by 4 has a remainder of 0 or 1, never 3. Therefore, it doesn't exist such n that the expression above is a perfect square.