$x^n + y^n = c$ has finitely many integral solutions? Assume $n > 1$ and $n$ is odd because it's easy if $n$ is even. 
Please help prove this.
$x^n + y^n = c$ has finitely many integral solutions if $c \neq 0$?
Thank you all for replying. I think I've just found my answer to this one. And I think it's quite simple. Here it goes.
(Non-negative solutions mean solutions $(x,y)$ where $x, y \geq 0;$ and negative solutions mean solutions $(x, y)$ where $x, y < 0$.)
The equation has finitely many non-negative solutions as well as negative solutions. This is quite trivial to show. From here, consider only the solutions $(x, y)$ where $x$ and $y$ have opposite signs. Without loss of generality, assume $x >0$ and $y <0$.
Let $z = -y >0$. So the equation becomes $x^n - z^n = c$.
$\implies |x^n - z^n| = |x-z||x^{n-1}+x^{n-2}z + \dots+xz^{n-2}+z^{n-1}|=|c|$.
Since $|x-z| \geq 1$ (if $|x-z| = 0, c = 0$), $|c| \geq |x^{n-1}+x^{n-2}z + \dots+xz^{n-2}+z^{n-1}| \geq |x^{n-1} + z^{n-1}| = x^{n-1} +z^{n-1}$
Now, it's obvious that there are only finitely many such $(x, z)$. And it's pretty much done here. Please let me know of my any mistake.

Let's look at the factorization of $x^n+y^n$.
For all odd $n$, $$x^n+y^n=(x+y)\left(x^{n-1}-x^{n-2}y+x^{n-3}y^2-x^{n-4}y^3+\cdots-xy^{n-2}+y^{n-1}\right)$$
Thus, since $x^n+y^n=c$, $$\left(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1} \right)\Bigg| c$$
This implies that $$\left|x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}\right|\leq |c|$$
Now here's the key to the argument: we claim that $$x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}\geq \min\left(x^{n-1}, y^{n-1} \right)$$
If we can show this, then we will have shown that in order for $x^n+y^n=c$, we must have that $$\min\left(x^{n-1},y^{n-1}\right)\leq |c|$$
This wins the proof for us because this, combined with our restriction that $x^n+y^n=c$, is only true for finitely many $x$ and $y$. So all we have to do now is show the inequality holds. I'll do the $n=5$ case so you can see the general principal at work.
So we're considering the equation $$x^4-x^3y+x^2y^2-xy^3+y^4$$
If $|x|=|y|$ (this implies that $x=y$ since $x\neq -y$ because $x^n+(-x)^n=0$ and $c\neq 0$), then it's very easy to show that the inequality holds. $\checkmark$ 
Without loss of generality, let's let $|x|>|y|$. We're gonna get a few cases.
Case 1 $x\geq 0$ and $y\geq 0$. Then we have that $x^4-x^3y\geq 0$ and $x^2y^2-xy^3\geq 0$, both since $x>y$. Thus $$x^4-x^3y+x^2y^2-xy^3+y^4\geq y^4= \min\left(x^4,y^4\right) \checkmark$$
Case 2 $x\leq 0$ and $y\leq 0$. This case is extremely similar to Case 1. $\checkmark$
Case 3 $x\geq 0$ and $y\leq 0$. Then we have that $-x^3y\geq 0$ and $-xy^3\geq 0$, both since $y\leq 0$. Thus $$x^4-x^3y+x^2y^2-xy^3+y^4\geq y^4=\min\left(x^4,y^4\right) \checkmark$$
Case 4 $x\leq 0$ and $y\geq 0$. Super similar to Case 3. $\checkmark$
These are all of the possible cases, so that completes the proof for $n=5$. (Woohoo!)
Showing that that inequality holds for all odd $n$ is done using the same general idea as in the $n=5$ case. Then, once you've shown that, there are only finitely many $x$ and $y$ that can satisfy $$\min\left( x^{n-1},y^{n-1}\right)\leq |c|$$ and $$x^n+y^n=c$$ so we're finished!
