Find all primes $(p,q)$ satisfying the condition that $pq$ divides $2^p+2^q.$ 
Find all primes $(p,q)$ satisfying the condition that $pq$ divides $2^p+2^q.$

I get answers for $p =2$ or $q = 2$ but I want to know the general approach for this question.

Assume both $p$ and $q$ are primes at least 3. Let $a_p$ denote the smallest integer satisfying $2^{a_p} \equiv_p -1$ and let $a_q$ denote the smallest integer satisfying $2^{a_q} \equiv_q -1$. [There indeed exists such an $a_p,a_q$ if there exists a $p,q$ satisfying the equation $pq|(2^p+2^q)$; indeed [assuming $q \ge p$]
$$pq|(2^p+2^q) \quad \Rightarrow \quad pq|(1+2^{q-p})$$ $$\Rightarrow \quad 2^{q-p} \equiv_p -1; \ 2^{q-p} \equiv_q -1.$$ Thus given $2^{q-p} \equiv_p -1$ and $2^{q-p} \equiv_q -1$ there indeed exists such an $a_p,a_q$ as claimed.]
Also note that $2^{p} \equiv_p 2$ and $2^{q} \equiv_q 2$. Then  $$pq|(2^p+2^q); \ 2^p \equiv_p 2; \ 2^q \equiv_q 2 \quad \Rightarrow \quad 2^{q} \equiv_p -2; \ 2^p \equiv_q -2$$
$$\Rightarrow \quad 2^{q-1} \equiv_p -1; \ 2^{p-1} \equiv_q -1.$$ As both $$2^{q-1} \equiv_p -1$$ and $$2^{q-1} \equiv_q 1$$ hold then, it follows that $q-1 = \ell a_q$ for some even integer $\ell$ and $q-1 = ka_p$ for some odd integer $k$. Thus what follows is the strict inequality $$\nu_2(a_p) > \nu_2(a_q),$$ where $\nu_2(M)$ for each integer $M$ is defined as the largest power of $2$ dividing $M$.
Likewise, $$2^{p-1} \equiv_q -1.$$ And, $$2^{p-1} \equiv_p 1.$$ So it follows that $p-1 = b a_p$ for some even integer $b$ and $p-1 = ca_q$ for some odd integer $c$. Thus what also follows is the strict inequality: $$\nu_2(a_p) < \nu_2(a_q).$$
However, the strict inequalities $\nu_2(a_p) > \nu_2(a_q)$ and $\nu_2(a_p)<\nu_2(a_q)$ contradict each other, so indeed, there are no two odd primes $p,q \ge 3$ satisfying  $pq|(2^p+2^q)$. Thus the result follows at least for any two primes $p,q \ge 3.$

If $p=2$ then $$2^p+2^q \equiv_q 4(1+2^{q-2}) \equiv_q 4(1+2^{-1}),$$ where $2^{-1}$ is defined to be the multiplicative inverse of $2$ in $\mathbb{Z}/q\mathbb{Z}$. [Indeed as $2^{q-1} \equiv_q 1$ it follows that $2^{q-2} \equiv_q 2^{q-1}2^{-1} \equiv_q 1$]. So for $q$ to divide $2^{p}+2^{q}$ for $p=2$ it follows that either the equation $4\equiv_q 0$ or the equation
$1+2^{-1} \equiv_q 0$ holds. The former equation gives $q=2$. The latter equation gives $q=3$. [Indeed, $2^{-1} \equiv_q -1$ which gives $2 \times 2^{-1} \equiv_q 2 \times -1$ which gives $1 \equiv_q -2$ which indeed gives $q=3$.] So if $p$ is $2$ then $q$ must be $2$ or $3$. However, note that $p=2,q=2$ and $p=2,q=3$ are both solutions.
