648. Find all natural $n$, for which the number $2^{n}+65$ is a perfect square.
$\triangle$ Let's consider two cases.

1) Suppose $n$ is even: $n=2 k(k \in N)$. Then

$$
2^{2 k}+65=a^{2},
$$

where $a$ is a natural number. We get:

$$
a^{2}-2^{2 k}=65, \quad\left(a+2^{k}\right)\left(a-2^{k}\right)=65
$$

When is the product of two natural numbers equal to 65? There are two possibilities:

$$
\left\{\begin{array} { l } 
{ a + 2 ^ { k } = 6 5 , } \\
{ a - 2 ^ { k } = 1 }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
a+2^{k}=13 \\
a-2^{k}=5
\end{array}\right.\right.
$$

From these systems of equations, we find $2^{k}$, then $-k$ and finally $n=2 k$. We get that $n=10$ or $n=4$.

2) Suppose $n$ is odd.

Then the power $2^{k}$ ends in the digit 2 or 8 (check this), which means the sum $2^{n}+65$ ends in the digit 7 or 3. But a perfect square cannot end in 7 or 3, so all odd $n$ must be discarded.

Answer: $10,4$.