$ (x\!-\!z_1)(x\!-\!z_2) = x^2\! - (z_1\!+\!z_2) x + z_1 z_2\in \Bbb R[x]$ has discriminant $\,(z_1\!+\!z_2)^2\!-4z_1z_2 > 0,$ since $z_1z_2 < 0,$ hence the roots $z_1,z_2$ are real (by the quadratic formula).
Remark $\ $ The proof works more generally for $\ z_1,z_2$ elements of any (integral) domain $\,\Bbb D\supset \Bbb R,$ assuming only that $\,z_1\!+z_2,\, z_1 z_2\,$ are both in $\Bbb R,$ and $\,z_1 z_2 < 0$. The hypothesis that $\Bbb D $ is a domain ensures that the list of roots $z_1,z_2 \in \Bbb R$ obtained by the quadratic formula persists as the unique list of roots in $\Bbb D$ (if the quadratic polynomial had an additional root $z_3\in \Bbb D$ then it would have more roots than its degree, which cannot occur in a domain). The claim may fail in extension rings not domains, e.g. $ f(x) = (x+2)(x-1) = x^2+x-2 = (x-w)(x+w+1)\, $ over $\,\Bbb R[w]/(w^2+w-2),\,$ where $f$ has non-real roots $\ w,\, -w-1 \not\in \Bbb R,\,$ contra to the claim.
Notice that said extension ring is not a domain since there $\,(w+2)(w-1) = 0,\,$ but $\,w \ne -2,1$. Informally, one may think of $w$ as an algebraic representation of a "generic" root of $f$, sharing all $(\Bbb R -)$algebraic properties of the roots $-2,1,$ i.e. $w$ satisfies $f(w) = g(w)$ for polynomials $f,g\in\Bbb R[x]\,$ iff $\,-2$ and $1$ also do: $f(-2) = g(-2)$ and $f(1) = g(1).\,$ While we can always construct extension rings containing such generic roots, generally the construction does not preserve the property of being a domain, and the consequent uniqueness of the list/multiset of roots (which, suitably formulated, is a characteristic property of domains among rings). Yet another example that uniqueness theorems provide powerful tools for deducing equalities.