Prove that $\exists z\in \Bbb R\forall x\in \Bbb R^+\Bigr[\exists y\in\Bbb R(y-x=\frac{y}{x}) \text{iff}\ x\neq z\Bigr]$.

Question

Not a duplicate of

Prove that $\exists z \in \mathbb{R} \forall x \in \mathbb{R}^+ [\exists y \in \mathbb{R} (y-x = y/x) \iff x \neq z]$

Prove that $\exists z \in \mathbb R\forall x \in \mathbb R^+[\exists y \in \mathbb R(y-x=y/x) \iff x \neq z]$

This is exercise $3.4.13$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Prove that $\exists z\in \Bbb R\forall x\in \Bbb R^+\Bigr[\exists y\in\Bbb R(y-x=\frac{y}{x}) \text{iff}\ x\neq z\Bigr]$.

Here is my proof:

Suppose $z=1$. Let $x$ be an arbitrary element of $\Bbb R^+$.

$(\rightarrow)$ Suppose $\exists y\in\Bbb R(y-x=\frac{y}{x})$. So we can choose some $y_0$ such that $y_0-x=\frac{y_0}{x}$. Let $x=z=1$. Substituting $x=1$ into $y_0-x=\frac{y_0}{x}$ we obtain $y_0-1=y_0$ and ergo $-1=0$ which is a contradiction. Ergo $x\neq z$. Thus if $\exists y\in\Bbb R(y-x=\frac{y}{x})$ then $x\neq z$.

$(\leftarrow)$ Let $x\neq z=1$. Let $y=\frac{x^2}{x-1}$ which is defined since $x\neq1$. Then

$$y-x=\frac{x^2}{x-1}-x=\frac{x}{x-1}=\frac{y}{x}.$$

Thus if $x\neq z$ then $\exists y\in\Bbb R(y-x=\frac{y}{x})$.

Thus $\exists y\in\Bbb R(y-x=\frac{y}{x})\ \text{iff}\ x\neq z$. Since $x$ is arbitrary, $\forall x\in \Bbb R^+\Bigr[\exists y\in\Bbb R(y-x=\frac{y}{x})\ \text{iff}\ x\neq z\Bigr]$. Therefore $\exists z\in \Bbb R\forall x\in \Bbb R^+\Bigr[\exists y\in\Bbb R(y-x=\frac{y}{x}) \text{iff}\ x\neq z\Bigr]$. $Q.E.D.$

Is my proof valid$?$

Thanks for your attention.