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Fix positive integers $a,b$ and a nonzero integer $c$. In positive integer solutions of $a{x}^n-b{y}^n=c$ with $n\ge 1$ and $Y=\max (x,y)>1$, the exponent $n$ is bounded effectively in terms of $a,b,c$.

When $x\ne y$ and the larger-variable power is sufficiently large, put $\Lambda =\log (a/b)+n\log (x/y)$. The equation makes $\Lambda$ nonzero and givesA Baker lower bound for a homogeneous linear form in logarithms gives $\log |\Lambda |\ge -C(a,b)\log Y\log (2n)$: the algebraic numbers are the rationals $a/b$ and $x/y$, the latter has height at most $Y$, and the coefficients are $1,n$. Dividing by $\log Y\ge \log 2$ yields $n\le C\log (2n)+O(1)$, which bounds $n$. If the larger-variable power is bounded, so is $n$. For $x=y>1$, the equation $(a-b)x^n=c$ gives the bound directly.

This is a bound on the exponent. Fixed exponents one or two can still admit infinitely many pairs $(x,y)$; for instance, $2x^2-y^2=1$ is a negative Pell equation after exchanging variables. The Thue theorem must not be applied to this quadratic case. The constant solution $x=y=1$, possible when $a-b=c$, is excluded by $Y>1$.