Suppose that $x = \max\{x, y, z\}$. We see that if $x, y, z > 0$ then based on the AM-GM inequality, we have $x + y + z \ge 3\sqrt[3]{xyz} = 3$ and the equality must occur, which means that $x = y = z = 1$, is not satisfied.

Next, consider $y, z < 0$ and put $u = -y > 0$, $v = -z > 0$. One can rewrite the given conditions as $x = 3 + u + v$ and $xuv = 1$. Thus $x$ is always greater than the sum of the remaining two numbers, we need to find $k$ such that
$$\sqrt[k]{x} < \sqrt[k]{u} + \sqrt[k]{v}.$$
Checking with $u = v$, one can solve the given condition to get $(x, u, v) = (4, \frac{1}{2}, \frac{1}{2})$. Then, by substituting this tuple into the above inequality, one can get
$$\sqrt[6]{4} < \frac{2}{\sqrt[6]{2}} \implies \sqrt[6]{8} < 2.$$
From this we have $k > 3$ or $k \ge 4$. Now we will prove that $k = 4$ satisfies the given condition, that is,
$$\sqrt[4]{x} < \sqrt[4]{u} + \sqrt[4]{v} \iff \sqrt{x} < \sqrt{u} + \sqrt{v} + 2\sqrt[4]{uv} = \sqrt{u} + \sqrt{v} + \frac{2}{\sqrt[4]{x}}$$
We will prove the stronger inequality
$$\sqrt{x} < \sqrt{u} + \sqrt{v} + \frac{2}{\sqrt{x}} \iff \sqrt{x} - \frac{2}{\sqrt{x}} < \sqrt{u} + \sqrt{v}.$$
Since $x \ge 4$ then $\sqrt{x} - \frac{2}{\sqrt{x}} > 0$ and squaring both sides gives
$$x + \frac{4}{x} - 4 < u + v + 2\sqrt{uv} \iff \frac{4}{x} < 1 + \frac{2}{\sqrt{x}}.$$
The final inequality is true because $x > 3$. Therefore $k_{\min} = 4$. $\square$