Let's find the positive integer values of $n$ for which $2^{n}+65$ is a perfect square.
It is known that the last digits of the powers of 2 with positive integer exponents repeat periodically. The length of one period is 4, and the repeating digits are: $2,4,8,6$. If $n$ is odd, then $2^{n}$ ends in 2 or 8, and thus in the case of an odd exponent, the last digit of $2^{n}+65$ is 7 or 3. Since the last digit of a square number cannot be 3 or $7, n$ must certainly not be odd.

Now let $n=2k$, where $k$ is a positive integer. Can there be a positive integer $x$ such that:

$$
2^{2k}+65=x^{2}
$$

that is,

$$
65=x^{2}-\left(2^{k}\right)^{2}=\left(x-2^{k}\right)\left(x+2^{k}\right)
$$

Since $x+2^{k}$ and $x-2^{k}$ are positive integers, they must be divisors of 65. But $x-2^{k}<x+2^{k}$, so the following cases are possible:

$$
\begin{array}{ll}
x-2^{k}=1, & x+2^{k}=65 \\
x-2^{k}=5, & x+2^{k}=13
\end{array}
$$

Solving the two systems of equations, we get the following results:

$$
\begin{gathered}
x=33, \quad 2^{k}=32, \quad k=5, \quad n=10 \\
x=9, \quad 2^{k}=4, \quad k=2, \quad n=4
\end{gathered}
$$

Thus, $n=4$ and in this case $2^{n}+65=9^{2}$, or $n=10$ and in this case $2^{n}+65=33^{2}$.