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The abc conjecture says that for every $ϵ>0$ there is $C_{ϵ}$ such that coprime positive integers $A+B=C$ satisfy $C\le C_{ϵ}\mathrm{rad}(ABC{)}^{1+ϵ}$, where the radical is the product of distinct prime divisors.

The Catalan alternative yields finiteness of actual nontrivial solutions, with $x,y\ge 2$ and $p,q\ge 3$. Put $M=x^p$, so $y^q=M-1$ and $x,y$ are coprime. The triple $(y^q,1,x^p)$ has radical at most $xy<M^{2/3}$. Take $ϵ=1/6$ to obtainBoth bases are bounded, and $2^p\le M$, $2^q\le M-1$ also bound both exponents. Thus only finitely many nontrivial Catalan solutions exist under abc, the Catalan power bound from abc.

For the Fermat alternative, the usual arithmetic height statement concerns primitive solutions, or equivalently solutions modulo common scaling. A primitive positive triple is pairwise coprime, and $x,y<z$. Applying abc to $(x^n,y^n,z^n)$ givesThis is uniform even in $n$. If $h=\max (x,y)<z$, then $z^n=x^n+y^n\le 2h^n$, so $n\le \log 2/\log (z/h)$ is bounded after $z$ is bounded. There are therefore finitely many primitive triples and exponents, the primitive Fermat bound from abc. Without the primitive convention, any one nonzero homogeneous solution would produce infinitely many common multiples. The Catalan proof above supplies the requested literal finiteness alternative without that normalization issue.