Let me explain how to show that the projective surface $\Sigma$ is geometrically irreducible (see also the comments above).

First, we know that $\Sigma$ is irreducible (this was checked by the OP).

Also, a routine calculation should reveal that the point $(1:2:0:0:0)$ is nonsingular (the OP also checked this).

As the point (1:2:0:0:0) it is obviously $\mathbb{Q}$-rational, it follows from the following lemma that $\Sigma$ is geometrically irreducible (i.e., $\Sigma_{\mathbb{C}}$ is irreducible).

Lemma. Let $X$ be an irreducible finite type scheme over a field $k$. Suppose that there is a point $P\in X(k)$ such that $X$ is smooth at $P$. Then $X$ is geometrically irreducible.

Proof. Let $S\subset X$ be the smooth locus of $X$. Note that $S$ is a dense open subscheme. (It is always open. Density follows from the fact that it is empty and that $X$ is irreducible: any non-empty open of an irreducible scheme is dense.) Since $S$ is a dense open of an irreducible scheme, it is itself irreducible. Now, by assumption $S(k)\neq \emptyset$. Therefore, $S$ is geometrically connected. (To see this, use that Galois action of the absolute Galois group permutes the connected components of $S_{\overline{k}}$, but fixes $P$.) Since $S$ is smooth and geometrically connected, it is geometrically irreducible. It follows that $S_{\overline{k}}$ is an integral dense open subscheme of $X_{\overline{k}}$. Thus $X$ is geometrically irreducible. QED

Let me explain how to show that the projective surface $\Sigma$ is geometrically irreducible (see also the comments above).

First, we know that $\Sigma$ is irreducible (this was checked by the OP).

Also, a routine calculation should reveal that the point $(1:2:0:0:0)$ is nonsingular (the OP also checked this).

As the point (1:2:0:0:0) is obviously $\mathbb{Q}$-rational, it follows from the following lemma that $\Sigma$ is geometrically irreducible (i.e., $\Sigma_{\mathbb{C}}$ is irreducible).

Lemma. Let $X$ be an irreducible finite type scheme over a field $k$. Suppose that there is a point $P\in X(k)$ such that $X$ is smooth at $P$. Then $X$ is geometrically irreducible.

Proof. Let $S\subset X$ be the smooth locus of $X$. Note that $S$ is a dense open subscheme. (It is always open. Density follows from the fact that it is non-empty and that $X$ is irreducible: any non-empty open of an irreducible scheme is dense.) Since $S$ is a dense open of an irreducible scheme, it is itself irreducible. Now, by assumption $S(k)\neq \emptyset$. Therefore, $S$ is geometrically connected. (To see this, use that Galois action of the absolute Galois group permutes the connected components of $S_{\overline{k}}$, but fixes $P$.) Since $S$ is smooth and geometrically connected, it is geometrically irreducible. It follows that $S_{\overline{k}}$ is an integral dense open subscheme of $X_{\overline{k}}$. Thus $X$ is geometrically irreducible. QED