Effect of abc conjecture on Fermat's Last Theorem A website ( http://www.math.unicaen.fr/~nitaj/abc.html#Consequences ) says that the $abc$ conjecture implies that there are only finitely many solutions to the equation $x^n+y^n=z^n$ with $\gcd(x,y,z)=1$ and $n\ge 4$. This one I have proven.
Lang's Algebra (p. 196) says that the $abc$ conjecture implies that for all $n$ sufficiently large, there are no solutions to the equation $x^n+y^n=z^n$ with $x,y,z\ne 0$. This one I have not proven (is it true?).
What are the strongest known assertions about Fermat's Last Theorem that follows from the $abc$ conjecture, and how are they proven? I searched the web but people tend to just say vague things like "implies asymptotic version of FLT" and such.
 A: There is a slight ambiguity what "the ABC conjecture" is as there are some variation. However, the most common and what you likely mean is this fomulation (or something equialent to it): 
For every $\epsilon >0$, there is a $C_{\epsilon}$ such that: if $a+b=c$, with positive coprime intergers, then 
$$c < C_{\epsilon} \ \text{rad}(abc)^{1 +\epsilon}.$$ 
Now, for the equation $x^n + y^n = z^n$ this means that 
$$
z^n < C_{\epsilon}\ \text{rad}(x^ny^nz^n)^{1 +\epsilon}.
$$
Yet  $\text{rad}(x^ny^nz^n) = \text{rad}(xyz)$ so one actually has
$$
z^n < C_{\epsilon} \ \text{rad}(xyz)^{1 +\epsilon}.
$$
and since $z$ is the largest and since $\text{rad}(a) \le a$ this further means
$$
z^n < C_{\epsilon}  z^{3 + 3\epsilon}.
$$
Now, take $\epsilon =1/4$, say. Then on the one hand this cannot hold for any $z>1$ for $n$ sufficiently large (so no solution for large $n$, what you ask) and also not for any $n \ge 4$ fixed for $z$ sufficiently large (so only finitely many for fixed $n$) or also only finitely many couples $(z,n)$ that fulfill this (for $n \ge 4$).
[Added:] Since the question was changed to remove the gcd condition in Langs's version, I add for completeness, that each solution (for given $n$) implies the existence of a coprime one (for this $n$) so since above establishes there are no coprime solutions for some $n$ then there are none at all. [End Added] 
However, to make these things effective/explicit one would need to know something about  $C_{\epsilon}$ (in dependence of $\epsilon$).   
Regarding "strongest possible": I think this is about what can be said, from the conjecture in the way I stated it. If one assumes stronger conjectures where one would have an explicit dependence of $C_{\epsilon}$ on $\epsilon$ then one could give explicit bounds. But (I think) the argument essentially always passes throught the last displayed equation and checking what this yields.   
