This is an easy problem, wonder why no one solved it. We assume that there exists a solution $(x_0,y_0) $. Then clearly $(x_0,y_0)\in\mathbb {R}^+$. Multiplying the two equations, we get $\ln (x_0+y_0)\ln (x_0-y_0)=(\ln x_0)^2$. Define a function $f:\left (-x_0,x_0\right)\longrightarrow \mathbb {R}$ by setting $f (y)=\ln (x_0+y)\ln (x_0-y)\forall y\in\left(-x_0,x_0\right) $. Then we see that $y_0$ is a solution of $f (y)=f (0) $. We observe that $f'(y)=\frac {\ln (x_0-y)}{x_0+y}-\frac {\ln (x_0+y)}{x_0-y} $. By a simple inspection, $f'(y)>0\forall y\in\left (-x_0,0\right)$,$f'(y)<0\forall y\in\left (0,x_0\right)$,$f'(0)=0$. Hence $0$ is the global maxima in the domain of $f $,i.e., $f (0)>f (y)\forall y\in\left (-x_0,0\right)\cup\left (0,x_0\right)$, hence $f (y_0)=f (0)\implies y_0=0$, which contradicts $y_0\in\mathbb{R}^+$. This contradiction proves that there are no solutions.

It has more easier solution. First, $x>y>1$ Second $\lg{(x^2-y^2)}=\lg x \lg y+ \frac{\lg x}{\lg y} \geq 2\lg x = \lg(x^2)$ - contradiction

RagvaloD wrote: It has more easier solution. First, $x>y>1$ Second $\lg{(x^2-y^2)}=\lg x \lg y+ \frac{\lg x}{\lg y} \geq 2\lg x = \lg(x^2)$ - contradiction Very nice @RagvaloD sir. Also @babu2001 your solution was nice but extra in some sense.

@RagvaloD, why is $1 <y<x$?

babu2001 wrote: @RagvaloD, why is $1 <y<x$? Easy to infer $$x-y > 0$$so $x>y$ Now, If $y<1<x$ then, $\lg y <0 , \lg x >0 \implies \lg x \lg y <0$ , but $\lg(x+y) > 0$ If $y<x<1$ then $\lg (x-y) <0$ but $\frac{ \lg x}{ \lg y} > 0$