Let $k$ be a positive integer. We choose a prime divisor $p_j$ of $2^{2^j} + 1$ for each $0 \le j < k$. Then the product $n_k = p_0p_1\dots p_{k-1}$ is a divisor of $2^{2^k} - 1 = \prod_{j=0}^{k-1}(2^{2^j} + 1)$.

Since $n_k$ is odd, $\sigma(n_k) = (p_0+1)\dots(p_{k-1}+1)$ is divisible by $2^k$. Hence $2^{2^k}-1$ divides $2^{\sigma(n_k)}-1$. It follows that $2^{\sigma(n_k)} \equiv 1 \pmod{n_k}$.