<s>[INST] On the divisibility of $(x+y)^k - 1$ by $xy$ For a fixed $k \geq 2$, are there infinitely many non-trivial coprime integer pairs $(x,y)$ for which $xy$ divides $(x+y)^k-1$? By trivial I mean parametrized pairs $(x,y)$ of the form $(-1,-1),(x,1),(1,y),(x,1-x),(x,1 - x^n),(1-y^n,y),(x,p(x)),(p(y),y)$, where $p(x)$ is a polynomial factor of $\frac{x^k-1}{x-1}$. When $k$ is even there are a few more trivial pairs, namely $(x,-1),(-1,y),(x,-1-x),(x,x^n-1),(y^n-1,y)$; where $n>0$ is a multiple of $k$.
 [/INST] For fixed $k$ and $y=\frac{x^k-1}{x-1}$ we have $$x^k \equiv 1 \pmod y$$and $$y \equiv 1 \pmod x$$ so also $$y^k\equiv 1 \pmod x.$$
That isn't one of your trivial examples, but maybe it should be. 
Actually, $y$ could be any polynomial factor of $\frac{x^k-1}{x-1}.$ For example,
$\frac{x^{30}-1}{x-1}=\Phi_3\Phi_5\Phi_{15}\Phi_{30}=$
$ (x^2+x+1)(x^4+x^3+x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$$(x^8+x^7-x^5-x^4-x^3+x^2+1)$
So for $k=30$ one has $x,y$ where $y$ is any of the $16$ products of the factors.
Maybe you want a parametric family that is not $y=p(x).$ 
Interesting Variant To have $xy$ divide $(x+y)^2+1$ is quite interesting. I saw it in a movie (in another form). It may have been a math olympiad problem.
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