The first two equalities imply $x>m$ and $y>n$ so one can make substitutions $x=m+X$, $y=n+Y$ and $k=X+Y$, with still $X,Y \in \mathbb N$:

${X \choose 2}=nX+nY-mX\tag{1}$

${Y \choose 2}=mX+mY-nY\tag{2}$

From (1) follows: $\quad m=n+n\frac{Y}{X}-\frac{1}{X}{X \choose 2}$,

then eliminate $m$ from (2): $\quad {Y \choose 2}+{X \choose 2}+\frac{Y}{X}{X \choose 2}=nX+nY+n\frac{Y^2}{X}$,

and finally: $\quad 2n=X-\frac{X^2+2XY}{X^2+XY+Y^2}$.

Clearly $X-2n=\frac{X^2+2XY+0Y^2}{X^2+XY+Y^2}\in(0,2)$, but since it is an integer it can only be $1$.

This proves that $X=2n+1$ and by symmetry $Y=2m+1$. Substituting for $X$ and $Y$ in (1) or (2) above finally yields $m=n$, so $X=Y$ and $x=y$.

The first two equalities imply $x>m$ and $y>n$ so one can substitute $x=m+X$, $y=n+Y$ and $k=X+Y$, with still $X,Y \in \mathbb N$:

${X \choose 2}=nX+nY-mX\tag{1}$

${Y \choose 2}=mX+mY-nY\tag{2}$

From (1) follows: $\quad m=n+n\frac{Y}{X}-\frac{1}{X}{X \choose 2}$,

then eliminate $m$ from (2): $\quad {Y \choose 2}+{X \choose 2}+\frac{Y}{X}{X \choose 2}=nX+nY+n\frac{Y^2}{X}$,

and finally: $\quad 2n=X-\frac{X^2+2XY}{X^2+XY+Y^2}$.

Clearly $X-2n=\frac{X^2+2XY+0Y^2}{X^2+XY+Y^2}\in(0,2)$, but since it is an integer it can only be $1$.

This proves that $X=2n+1$ and by symmetry $Y=2m+1$. Substituting for $X$ and $Y$ in (1) or (2) finally yields $m=n$, so $X=Y$ and $x=y$.