For which positive integer values of $n$ will $2^{n}+1$ be a power of a positive integer greater than one?
If $2^{n}+1=a^{k}$, then the left side is odd, and thus $a$ is also odd. Subtracting 1 from both sides, the right side can be factored according to a known identity:

$$
2^{n}=a^{k}-1=(a-1)\left(a^{k-1}+a^{k-2}+\cdots+a+1\right)
$$

According to the resulting equation, we have expressed a power of 2 as the product of positive integers. By the fundamental theorem of arithmetic, this is only possible if the factors themselves are powers of 2. Since $a$ is odd, the second factor on the right side, being the sum of $k$ odd numbers, gives the same remainder when divided by 2 as $k$ does. This factor, by the condition, is greater than the only odd power of 2, which is 1, so if it is a power of 2, then $k$ must be even. Therefore, $a^{k}$ is a perfect square, $a^{\frac{k}{2}}=b$ is its square root.

Our equation then takes the significantly simpler form

$$
2^{n}+1=b^{2}
$$

where $b$ is odd. Factoring again,

$$
2^{n}=(b-1)(b+1)
$$

thus, according to our previous reasoning, both $b-1$ and $b+1$ are powers of 2. Since the difference of powers of 2 is divisible by the smaller one, $(b-1)$ is an even power of 2 that divides 2: this is only possible if $b-1=2$.

Then $b=3, b^{2}=9$, which is of the form $2^{3}+1$, giving us the (only) solution to the problem.

Thus, $2^{n}+1$ is a power of a positive integer greater than 1 with an integer exponent greater than 1 only in the case $n=3$.