Prove that there are no integer solutions to $x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$

Rewriting the equation as $y/x=x(y^2-1)-2y$, we see that we must have $x\mid y$ (since the right hand side is an integer). So letting $y=xu$ (with $x\not=0$), we get

$$u=x(x^2u^2-1)-2xu$$

which implies $x\mid u$ and $u\mid x$, so $u=\sigma x$ with $\sigma=\pm1$. But this gives

$$\sigma x=x(x^4-1)-2\sigma x^2$$

which simplifies (on cancelling out an $x$) to

$$x^4-2\sigma x-1-\sigma=0$$

and neither $x^4-2x-2=0$ nor $x^4+2x=0$ has any (nonzero) integer roots.

Well you can't have $x=0$ so multiply through by $x$ to obtain $$x^2(y^2-1)=y(2x+1)$$

Then either you have $y=\pm 1$ [or $y=0$] (which you can exclude) or the left-hand side is positive.

Now compare the terms in $x$ on either side (careful that $2x+1$ may be negative) and the terms in $y$ on either side (with similar care).

We are given $x(y^{2}-1)=y\left(2+\dfrac{1}{x}\right)$ with $x,y\in\Bbb Z$. The presence of the $1/x$ term implies $x\neq0$ and hence $y\neq0$. Multiplying through by $x$ gives $$x^2(y^2-1)=y(2x+1).$$Note that $2x+1$ is odd. Hence $y$ cannot be odd, because then $y^2-1$ would be even, and our equation would equate an even to an odd number. So $y$ is even. Hence $x^2$ is even, and therefore so is $x$. It follows that $y$ is divisible by $4$. Then $|(y^2-1)/y|=|y-1/y|>3$, while $|(2x+1)/x^2|=|2/x+1/x^2|<2$. Consequently our equation cannot be satisfied.