@user193319 The wlog assumption is valid in both (2.1) and (2.2). Write $[x,y]=x^{bp}$; this is possible since $[x,y]\in\langle x\rangle$ and $|x|=p^2$. Also, $b$ is invertible mod. $p$, because $[x,y]\ne 1$. Thus, there exists a multiplicative inverse $c$ of $b$ mod. $p$. Now, because $x$ and $y$ commute with $[x,y]$, we have $[x,y^c]=[x,y]^c$. Furthermore, $[x,y]^c=x^p$. Replacing $y$ with $y^c$, we get the desired equality.