= Solution
The <abc conjecture> says that for every $\epsilon>0$ there is $C_\epsilon$ such that <coprime> positive <integers> $A+B=C$ satisfy $C\le C_\epsilon\operatorname{rad}(ABC)^{1+\epsilon}$, where the radical is the product of distinct prime <divisors>.
The Catalan alternative yields finiteness of actual nontrivial solutions, with $x,y\ge2$ and $p,q\ge3$. 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 $\epsilon=1/6$ to obtain
$$
M\le C_{1/6}M^{7/9},\qquad\boxed{M\le C_{1/6}^{9/2}}.
$$
Both bases are bounded, and $2^p\le M$, $2^q\le M-1$ also bound both exponents. Thus \b[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 $(x^n,y^n,z^n)$ gives
$$
z^n\le C_{1/6}(xyz)^{7/6}<C_{1/6}z^{7/2},\qquad\boxed{z\le C_{1/6}^2\quad(n\ge4)}.
$$
This is uniform even in $n$. If $h=\max(x,y)<z$, then $z^n=x^n+y^n\le2h^n$, so $n\le\log2/\log(z/h)$ 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.
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