 Solution

In the affine chart $z=1$, the equation is $y^2=x^2(x+1)$, irreducible because $x+1$ is not a square in $k(x)$. The partial derivatives show the unique singular point is $P=[0:0:1]$; the point at infinity is smooth.

A line $y=tx$ through $P$ meets the curve again atThus $t\mapsto [t^2-1:t(t^2-1):1]$ is birational, with inverse $t=y/x$ away from $P$. The incidence closure has one point over every smooth $Q$ and two points over $P$, corresponding to $t=\pm 1$. This is the normalization of a nodal curve and separates the two branches.

Dominated by projective space does not imply being projective space. A smooth quadric surface $Y\simeq {\mathbb{P}}^1\times {\mathbb{P}}^1$ is rational and hence dominated by ${\mathbb{P}}^2$, but is not any ${\mathbb{P}}^k$; for example its Picard rank is $2$, not $1$.