By plugging $x \to x+1$ we get $2xy=2yx,$ so if $2 \neq 0,$ we are done. Now let's assume $2=0.$ We have $(xy)^2y=y(xy)^2,$ so $(xy)^2y=(yx)^2y,$ implying $(xy)^2=(yx)^2.$ Now $$(xy-yx)^2=(xy)^2-xyyx-yxxy+(yx)^2=0,$$because $(xy)^2+(yx)^2=0$ and $xyyx+yxxy=yyxx+xxyy=y^2x^2+x^2y^2=0.$ Therefore $xy-yx=0.$ Remark. There is no need for $\mathbb{K}$ to be a division ring. It is enough do be a domain (i.e. ring without non-zero zero divisors).

Seventy four years ago, Kaplansky proved that if $K$ is a division ring such that for every $x \in K$ there exists a positive integer $n(x)$ such that $x^{n(x)}y=yx^{n(x)}$ for all $y \in K,$ then $K$ is commutative.

We first prove that $a$ and $-a$ commute. This is easy: \[ a^2+a(-a)=a(a-a)=0=(a-a)a=a^2+(-a)a\implies a(-a)=(-a)a. \]Clearly $0$ commutes with everything. Let $a$ and $b$ be nonzero elements such that $a+b\neq 0$. Then \[ (a+b)^2a=a(a+b)^2\implies a^3+aba+ba^2+b^2a=a^3+a^2b+aba+ab^2\implies ba(a+b)=ab(a+b)\implies ba=ab, \]as desired. Thus $\mathbb{K}$ is a field. $\square$