Main Question
What are the divisors of $q^{q+1} - 1$ for $q$ a prime congruent to $1$ modulo $4$, other than $q - 1$?
My Attempt
Since $q$ is odd, $q+1$ is even so that $q^{q+1} - 1$ is a difference of two squares. Therefore, we obtain $$q^{q+1} - 1 = \bigg(q^{(q+1)/2} + 1\bigg)\cdot\bigg(q^{(q+1)/2} - 1\bigg).$$ I conjecture that $$(q+1) \mid (q^{q+1} - 1),$$ but I do not see how to prove it.
Added March 04 2017 (Manila time)
Following ajotatxe's lead, if $q + 2$ is prime, I have $$q^{q + 1} - 1 \equiv (-2)^{\phi(q + 2)} - 1 \equiv 1 - 1 = 0 \pmod {q + 2}$$ by Euler's formula involving the totient function $\phi$, since $\gcd(-2, q + 2) = 1$ (because $q$ is an odd prime and $q+2$ is prime).
Therefore, $$(q + 2) \mid (q^{q+1} - 1)$$ if $q + 2$ is prime.
Is this correct?
Since $q+1$ is even, $$q^{q+1}-1\equiv (-1)^{q+1}-1= 1-1=0\pmod{q+1}$$