37th Iranian Mathematical Olympiad

We are going to construct infinite pairs like $(a,b)$ that suit the problem's condition. Consider $c$ be an arbitrary natural number then put $a=c^{n+1398}$ and $b=c^{m+2019}$. Then $$a^m+b^n=c^{mn}(c^{1398m}+c^{2019n}),$$ and $$a^{2019}+b^{1398}=c^{1398\times 2019}(c^{1398m}+c^{2019n}).$$ So obviously set of the prime divisors of $a^m+b^n$ and $a^{2019}+b^{1398}$ are the same. Since $c$ has been arbitrarily chosen we have infinite choices for $(a,b)$.