Here is a simple argument for the second question. I renamed the field $k$ to $K$ for clarity.
Observe that $K/\mathbb{Q}$ is a Galois extension, K^{\mathrm{Aut}(K/\mathbb{Q})}=\mathbb{Q}$, because$K$is algebraically closed. For any$g\in\mathrm{Gal}(K/\mathbb{Q})$g\in\mathrm{Aut}(K/\mathbb{Q})$ and any $x\in X$ we have $x^g\in X$, since $$P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq l\leq k,$$ using that the coefficients of $P_l$ are rational. This shows that the polynomials $$Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n,$$ are invariant under $\mathrm{Gal}(K/\mathbb{Q})$, \mathrm{Aut}(K/\mathbb{Q})$, i.e. they also have rational coefficients. It follows that the coordinates$z_{ij}\in K$are algebraic over$\mathbb{Q}$. P.S. Thanks to Qiaochu Yuan and Kevin Ventullo. 3 added 36 characters in body; added 87 characters in body Here is a simple argument for the second question. I renamed the field$k$to$K$for clarity. Observe that$K/\mathbb{Q}$is a Galois extension, because$K$is algebraically closed. For any$g\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$g\in\mathrm{Gal}(K/\mathbb{Q})$ and any $x\in X$ we have $x^g\in X$, because since $$P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq l\leq k,$$ using that the coefficients of $P_l$ are rational. This shows that the polynomials $$Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n,$$ are invariant under $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, \mathrm{Gal}(K/\mathbb{Q})$, i.e. they also have rational coefficients. It follows that the points coordinates$z_{ij}$z_{ij}\in K$ are algebraic .over $\mathbb{Q}$.
Here is a simple argument for 2): Every element of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ fixes the polynomials second question.
For any $P_l$, hence g\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$and any$\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$permutes the finite set x\in X$ we have $X$. It follows that for each x^g\in X$, because $$P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq j\leq n l\leq k,$$ using that the polynomial coefficients of$Q_j(z):=\prod_{i=1}^m(z-z_{ij})$has P_l$ are rationalcoefficients. This shows that the polynomials $$Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n, because it is$$ are invariant under $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. This shows \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, i.e. they have rational coefficients. It follows that each the points$z_{ij}\$ is are algebraic.