I recently tried to show that there are no integer solutions to the equation $x^4+ 131 = 3y^4$. After stumbling with some algebraic attempts I reached the conclusion that all forth powers (except of course powers of multiples of 5) are congruent to $1$ (mod $5$). Then we can write $x^4+ 1 = 3y^4$ (mod $5$), and because $x^4 $ and $y^4$ are either $0$ or $1$ there are no integer solutions.
Despite being able to solve it, I had no method beyond blindly aplying diferent modulos until one worked. Is there a way to calculate the modulos for which $n$th powers have but a few values? And more generally, is there a method for knowing how to more efficiently find an appropriate modulo to solve this kind of diophantine equations?
By Fermat's little theorem, for every prime number $p$ and every integer $x$ coprime to $p$ we have $$x^{p-1}\equiv1\pmod{p}.$$ So when considering a Diophantine equation with $n$-th powers, it makes sense to consider primes for which $p-1$ is a multiple of $n$. Or in other words, primes $p\equiv1\pmod{n}$.
There is of course no guarantee that reducing modulo any such a prime will lead to a contradiction (even when there are no integral solutions!), so it is still a matter of trial and error. But at least this narrows down the search a bit.