Now $p(p^2-1)=q(q^6-1)$. If $p\leq q$ then $p^2-1\geq q^6-1$ then $q^3\leq p\leq q$ then $q^2\leq 1<2^2\leq q^2$, which is false. Hence $p>q$ so $p^3=q^7+p-q>q^7$ so $p>q^2$ and $p>q^2q^{\frac{1}{3}}$ $p|q(q-1)(q+1)(q^2-q+1)(q^2+q+1)$ Now $q-1<p$ so $p\not|q-1$ Also $q<p$ so $p\not|q$ Also $q+1=q^2+1-q(q-1)\leq q^2+1-2\cdot 1<q^2<p$ so $p\not|q+1$ Also $q^2-q+1\leq q^2-2+1<q^2<p$ so $p\not|q^2-q+1$ Hence $p|q^2+q+1$ so $p\leq q^2+q+1$ $2|q(q+1)$ so $2\not|q^2+q+1$ If $p<q^2+q+1$ then $3p\leq q^2+q+1<3q^2$ then $p<q^2<p$, which is false. So $p=q^2+q+1$ From here I will show two ways to complete the solution. $q^2q^{\frac{1}{3}}<p=q^2+q+1\leq q^2\left(1+\frac{1}{2}+\frac{1}{4}\right)$ Then $q<\left(\frac{7}{4}\right)^3<6$ so $q=2$ or $q=3$ or $q=5$ When $q=2$, $p=q^2+q+1=7$ but $7^3-7=336>126=2^7-2$ When $q=3$, $p=q^2+q+1=13$ and $13^3-13=2184=3^7-3$ When $q=5$, $p=q^2+q+1=31$ but $p^3-p\equiv 3^3-3\equiv 3\mod 7$ and $q^7-q\equiv 0\mod 7$, by Fermat's Little Theorem. So $(p, q)=( 13, 3)$ is the only solution Alternatively: $q(q-1)(q+1)(q^2-q+1)(q^2+q+1)=p(p-1)(p+1)=q(q+1)(q^2+q+1)(q^2+q+2)$ So $(q-1)(q^2-q+1)=q^2+q+2$ since $q(q+1)(q^2+q+1)>0$ So $q^3-3q^2+q-3=0$ So $(q-3)(q^2+1)=0$ So $q=3$ since $q^2+1>0$ So $p=3^2+3+1=13$ $13^3-13=2184=3^7-3$ so $(p, q)=(13, 3)$ is the only solution.

nice solution!!

$(p,q)=(13,3) $

$p(p-1)(p+1)=q(q-1)(q+1)(q^2-q+1)(q^2+q+1)$ CLAIM1: $p>q^2$ PROOF: Assume opposite $p<q^2$ then, $q^6-p>q^7-q$ which is easyly contradicition. So from CLAIM1 we get $p =(q^2+q+1)$ $(q^2+q+1)(q^2+q)(q^2+q+2)=q(q-1)(q+1)(q^2-q+1)(q^2+q+1)$ $q^2+q+2=(q-1)(q^2-q+1)$ By mod 3 we get $q=3$ $p=13$ indeed solutions.

another solution here

We have $q^6 < q^7 - q = p^3 - p < p^3 \Longrightarrow p > q^2$, so after factoring the equation as \[p(p-1)(p+1) = q(q-1)(q+1)(q^2-q+1)(q^2+q+1)\]we get that $p \mid q^2+q+1$, but $p>q^2>\frac{1}{2}(q^2+q+1)$, so $p = q^2 + q + 1$. Substituting this back into the equation leads to the cubic $(q-3)(q^2+1) =0$, so $(p,q) = (13,3)$ is the only solution.