Answer: $n$ is prime or a power of two (including $n=1$).
Indeed, if $n$ is an odd prime, then the only divisor is $d=1$. Choose $a=2^{k}-2$ with $3 \leq k \leq n+1$, for example $a=6$. Then we need to check that $2^{k}-2+2=2^{k}$ is a divisor of $\left(2^{k}-2\right)^{n}-2^{n}$. This is the difference between two terms that both have exactly $n$ factors of 2, so the difference has at least $n+1$ factors of 2.

In the second case, we have $n=2^{m}$ with $m \geq 0$. If $m=0$, then there are no positive divisors $d \neq n$, so the condition holds because the requirement is vacuously true. Note also that $m=1$ gives the even prime 2. For all divisors $d \neq n$, we now observe that

$$a^{n}-2^{n}=a^{d e}-2^{d e} \equiv\left(-2^{d}\right)^{e}-2^{d e}=2^{n}\left((-1)^{e}-1\right) \quad \bmod a^{d}+2^{d}$$

is zero for all $a$. Thus, $a^{n}-2^{n}$ is a multiple of $a^{d}+2^{d}$, or in other words, $a^{d}+2^{d} \mid a^{n}-2^{n}$. This checks both cases.

Now suppose that $n$ is neither a prime nor a power of two. Then we can write $n$ as $n=d e$ with $e \neq 1$ odd (since $n$ is not a power of two) and $d \neq 1$ (since $n$ is not a prime). Since $(-1)^{e}-1=2$, the above calculation now implies that $a^{d}+2^{d} \mid 2^{n+1}$. This means that $a^{d}+2^{d}$ is a power of two, so $a^{d}=2^{k}-2^{d}$ for some $k$ implies $\left(\frac{a}{2}\right)^{d}=2^{k-d}-1$ but this implies that $k-d \geq 2$. Thus, this square is -1 modulo 4, which is impossible. If $d$ is odd, then we observe that

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

Since $d \neq 1$, $\frac{a}{2}+1$ is strictly less than $\left(\frac{a}{2}\right)^{d}+1$. The second factor in this product is a sum of an even number of terms with the parity of $\frac{a}{2}$ and a 1, so this factor is odd. Since the second factor is also greater than 1, this contradicts the fact that it must be a divisor of the power of two $2^{k-d}$.