Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$. 
Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.

We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?

Continuing your idea, let $a = 1+ \dfrac 1t, t > 0.$ Then, you can write your inequality as $xy-x-y > e^2-2e$, which is in terms of the variable $t:$
$$f(t) = t\ln\left(1+\frac 1t\right) + \ln\left(\left(1+\frac 1t\right)^{t+1}-2-\frac 1t\right) > 1+\ln(e-2).$$
Wolfram says that this is indeed a decreasing function in $t$, so the infimum will be:
$$\lim\limits_{t\to\infty} f(t) = 1+ \ln(e-2).$$
Then, the only thing to show now is the fact that $f(t)$ is decreasing, but I suspect this will take an ugly computation.
EDIT: Actually, the derivative did not turn out to be too bad:
\begin{align}f'(t) &= \\
&\dfrac{\dfrac{1}{t^2} + \left(1+\dfrac 1t\right)^{t+1}{\left(2\ln\left(1+\frac 1t\right)-\dfrac 1t-\dfrac{1}{t+1}\right)} + \left(2+\dfrac{1}{t}\right)
\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\left(1+\dfrac 1t\right)^{t+1} - \dfrac 1t-2} &
\\<0\end{align}
Finally, the above is equivalent to:
$$\left(1+\dfrac 1t\right)^{t+1} > \dfrac{\dfrac{1}{t^2} + \left(2+\dfrac 1t\right)\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\dfrac{1}{t}+\dfrac{1}{1+t} - 2\ln\left(1+\dfrac{1}{t}\right)},$$
which I think should be doable by brute-force expanding with a Taylor's series etc.
