As the title suggests, let $g$ be a primitive root mod $p$. Suppose $5 \equiv g^a \pmod{p}$ and $7 \equiv g^b \pmod{p}$, where $a,b<p-1$ (note that $(5,p)=(7,p)=1$ and $a,b\neq p-1$). Then $5^b \equiv 7^a \pmod{p}$. If $a<b$, then choose $(m,n)=(p-1-b,a)$, meanwhile if $a>b$, then choose $(m,n)=(b,p-1-a)$. It is easy to see that they work.