Solution:
Answer: $-2 \pm \sqrt{5}$
From $x^{2}+4 x \in \mathbb{Q}$, we deduce that $(x+2)^{2}=x^{2}+4 x+4$ is also rational, and hence $x=-2 \pm \sqrt{y}$, where $y$ is rational. Then $x^{3}-17 x=(26-6 y) \pm (y-5) \sqrt{y}$, which forces $y$ to be $5$. Hence $x=-2 \pm \sqrt{5}$. It is easy to check that both values satisfy the problem conditions.