obviusly $ q$ is odd. case(1):$ p = 2 \Rightarrow2^{q + 1} = q^2 + 7$ but for $ k\geq4$ we have $ 2^{k + 1} > k^2 + 7$(one can prove it by induction) therefore $ q < 4$ and $ q$ is odd prime,therefore $ q = 3$.obviusly $ (p,q) = (2,3)$ is a solution. case(2):$ p\geq3$ lemma.if $ a\geq{b}\geq{e}$ then $ b^a\geq{a^b}$. proof:let $ f(x) = \frac {x}{\ln{x}}$ then $ f^{\prime}(x) = \frac {\ln{x} - 1}{(\ln{x})^2}$,therefore on interval $ [e,\infty)$ we have $ f^{\prime}(x)\geq0$ and therefore in this interval $ f$ is increasing.... $ a\geq{b}\geq{e}\Rightarrow\frac {a}{\ln{a}}\geq\frac {b}{\ln{b}}\Rightarrow a\ln{b}\geq{b\ln{a}}\Rightarrow\ln{b^a}\geq\ln{a^b}\Rightarrow{b^a\geq{a^b}}$. from original equation by fermat little theorem we get that $ p|q + 7$ but $ p = q + 7$ is impossible, because $ p,q$ are odd.therefore$ p\leq\frac {q + 7}{2}$ if $ q\geq{p}$ then $ q\geq{p}\geq3 > e$ therefore $ p^q - q^p\geq0$ and now from original equation we get that $ p^q\leq7$which is impossible ,because $ p,q\geq3\Rightarrow{p^q\geq27}$ therefore $ p > q$ and also we have $ p\leq\frac {q + 7}{2}$ therefore $ q\leq\frac {q + 7}{2}\Rightarrow{q < 7}$ if $ q = 5$ because $ q < p\leq\frac {q + 7}{2}$ we have $ 5 < p\leq6$ which is impossible... if $ q = 3$ because $ q < p\leq\frac {q + 7}{2}$ we have $ 3 < p\leq5$ therefore $ p = 5$ and $ (p,q) = (5,3)$ satisfy original equation...therefore all solution are $ (p,q) = (2.3),(3,5)$

ali666 wrote: lemma.if $ a\geq{b}\geq{e}$ then $ b^a\geq{a^b}$. is there any solution,without using this inequality?

Yes it can be done by an simple solution as following . The first easy to check that $ p,q$ are different from each other . Apply Fermat's Theory we have $ 2p^q\equiv 2p (\mod q)$ $ p^q\equiv p (\mod q)$ Therefore :$ q|2p - 7$ and $ p|q + 7$ It is can be solve by inequalities . The solution of this problem are $ (p,q) = (2,3),(3,5)$

TTsphn wrote: $ p^q\equiv p (\mod q)$ Therefore :$ q|2p - 7$ and $ p|q + 7$ It is can be solve by inequalities . by inequalities,we obtain $ p = 2,q = 3$ or $ q = 2p - 7$...then?!how we can find solution of the form $ (p,2p - 7)$ ?

Anyone has an elementary proof?

wait for elementary proof ...

By TTsphn's post, $p=2,\, q=3$ or $q=2p-7$. (1) $p=5$ : $q=3$ and it is solution. (2) $p=7$ : $q=7$ and $2p^q-q^p=7^7$, contradiction. (3) $p>10$ : $2p^{2p-7}=(2p-7)^p+7<(2p)^p$ so $2^p>2p^{p-7}>2\cdot 8^{p-7}=2^{3p-20}$, contradiction. $\therefore (p,\, q)=(2,\, 3),\, (5,\, 3).$