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I'm not sure whether the second question is inadvertently misworded. As it is, the answer can easily be shown to be no. Fix $h$ and let $A$ be huge compared to $h$. Let $$ f(x,y)=A(x-y)(x-2y)\cdots (x-ny)+h y^n. $$ Then the Thue equation $f(x,y)=h$ has the $n$ solutions $(1,1),(2,1),\dots,(n,1)$. Here $h$ is very small compared to $\lVert f \rVert \ge n! A$. So we cannot expect a bound on the number of solutions independent of the degree of $f$.