650. Find all natural $k$ for which the number

$$
2^{k}+8 k+5
$$

is a perfect square.
$\triangle$ Checking shows that the value $k=1$ does not work, the value $k=2$ works, and the values $k=3,4,5,6$ do not work. It seems that no $k>2$ satisfies the condition of the problem. Let's prove our assumption.

Assume that there exists such a $k>2$ that the number $2^{k}+8 k+5$ is a perfect square. Since this number is odd, we have

$$
2^{k}+8 k+5=(2 a-1)^{2},
$$

where $a$ is an integer. We get:

$$
2^{k}+8 k+5=4 a^{2}-4 a+1, \quad 2^{k}+8 k+4=4 a(a-1) .
$$

In the last equation, the right-hand side is divisible not only by 4 but also by 8. However, the left-hand side is not divisible by 8, so such an equation is impossible. We have reached a contradiction. Therefore, no $k>2$ has the required property.

Answer: 2.