Irreducibility of Lemniscates

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I am reading S.Yu.Orevkov's new paper named "Irreducibility of Lemniscates", but I am stuck in the conclusion of this paper.

The main result of this paper:

Any lemniscate is an irreducible real algebraic curve in $\mathbb{C}$.

Orevkov says in the paper that this conclusion follows immediately from the Corollary in the paper.

Here is the Corollary:

Corollary 1. Let $P(z)$ be a polynomial in one variable with complex coefficients and let $f(x,y)$ be the real polynomial given by $$f(x,y)=P(x+iy)\overline{P}(x-iy)-1,$$ where $\overline{P}$ is the polynomial whose coefficients are the conjugates of those of $P$. Then, $f(x,y)$ is reducible over $\mathbb{C}$ if and only if $P(z)=P_{1}(z)^{d}$ for $d>1$ and some polynomial $P_{1}(z)$.

Here is what I have attempted:

Let $P(z)\in\mathbb{C}[z]$ where $z=x+iy$. Then, define $$f(x,y):=P(z)\overline{P(z)}-1=|P(z)|^{2}-1.$$

Set $\mathcal{A}:=\{z\in\mathbb{C}:|P(z)|=1\}$. Then, we have $$\mathcal{A}=\{z\in\mathbb{C}:|P(z)|^{2}=1\}=\{x,y\in\mathbb{R}:f(x,y)=0\}.$$

It then remains to show that $f(x,y)$ is irreducible over $\mathbb{C}$.

Suppose $f(x,y)$ is reducible over $\mathbb{C}$. Then, by the Corollary, $P(z)=P_{1}(z)^{d}$ for $d>1$ and some polynomial $P_{1}(z)$.

Then, $\mathcal{A}=\{z\in\mathbb{C}:|P_{1}(z)|^{d}=1\}$.

Then, I got stuck in finding a contradiction.

In the paper, Orevkov also explains that

indeed, since $\{|P|=1\}=\{|P|^{d}=1\}$, any lemniscate can be defined via a polynomial which is not a power of another.

But I don't understand how this explanation implies the conclusion of this paper.

Thank you for any hints or explanations!!