Maybe try using the following relationships of the roots and coefficients of a quadratic polynomial:
$\text{Let } a, b, c \in \mathbb{Q}.$
$$ax^2 + bx + c = a(x-x_1)(x-x_2) = a(x^2 -(x_1 + x_2) + x_1 \cdot x_2) = ax^2 - a(x_1 + x_2)x +a \cdot x_1 \cdot x_2.$$
From here we get the following system of equalities:
\begin{cases} x_1 + x_2 = -\frac{b}{a} \\\ x_1 \cdot x_2 = \frac{c}{a} \end{cases}
Notice that if $x_1 = x + \sqrt{z}$, then $x_1 + x_2 = x + \sqrt{z} + x_2 = -\frac{b}{a} \in \mathbb{Q}$, then we can solve for $x_2$:
$$x_2 = (-\frac{b}{a} - x) - \sqrt{z} = m - \sqrt{z}, \text{ where } m \in \mathbb{Q}$$
Now let's use the second relationship we proved:
$$x_1 \cdot x_2 = (x+\sqrt{z})(m-\sqrt{z}) = (xm - z) + (m - x)\sqrt{z} = \frac{c}{a} \in \mathbb{Q}$$
If $m - x \neq 0$ we could solve for $\sqrt{z}$ to get that it must be rational, but this contradicts our assumption that it is irrational, hence $m = x$, which implies $x_2 = m -\sqrt{z} = x - \sqrt{z}$.