Take the 2-minute tour ×  I suppose a conjecture implies this so
there might be an unconditional proof.  Let $F(x,y)=0$ be a
curve with infinitely many integral points $(u^m,v^n)$ where
$\gcd(u,v)=1$ infinitely often and $m \ge 3,n \ge 2$. Such
curves are easy to construct by starting with a parametrization
for example.  For a bivariate polynomial $F$ define
$\operatorname{High}(F)$ to be the sum of the highest degree
monomials (i.e., $\operatorname{High}(F) = F$ iff $F$ is
homogeneous and $\operatorname{High}(2x^3+3y^3+x+y)=2x^3+3y^3$).
Let $\gcd(\operatorname{High}(F),xy)=1$.  Under these conditions
is $\operatorname{High}(F)$ not square-free?   This can't be
relaxed to $m \ge 2 $  share|improve this question add comment
1 Answer  (Added some remarks from my comments)  Suppose that
$F=F_1F_2\cdots F_r$ with $F_i$ irreducible. If $F=0$ has
infinitely many points of the requested form, then so does one
of the factors. Furthermore, $\text{High}(F)=\text{High}(F_1)
\text{High}(F_2)\cdots\text{High}(F_r)$. So i