Consider the set of $(x,z)\in\mathbb R^2$ given by:
$x^{2i}\leq z$ with $x\in\mathbb R$
$x^{2i+1}\leq z$ with $x\in\mathbb R$
$x^{2i}\leq z$ with $x\in\mathbb R_{\geq0}$
$x^{2i+1}\leq z$ with $x\in\mathbb R_{\geq0}$
$x^{2i}\leq z$ with $x\in\mathbb R_{>0}$
$x^{2i+1}\leq z$ with $x\in\mathbb R_{>0}$
at $i\in\{1,2,\dots\}$?
Answers:
$\{(x, z) : x^{2k} \leq z\}$ is convex for all positive integers $k$. This follows from the fact that $(x, z) \mapsto x^{2k} - z$ is a convex function of $(x, z)$. One checks this easily by differentiating twice, yielding $2k(2k - 1)x^{2(k-1)}$, which is a nonnegative function.
$\{(x, z) : x^{2k + 1} \leq z\}$ is not convex. For example, $x = (0, 0)$ and $y = (-1, -1)$ are in this set for any $k$. Their average is $(-0.5, -0.5)$. But $(-0.5)^{2k +1} = -(0.5)^{2k + 1} > -0.5$, as soon as $k \geq 1$. So this set is not convex.
The following image depicts the situation. The blue region is this set for $k = 1$, and the red region is $k = 25$. You can see the counterexample clearly in this picture. The green and purple points are in both sets, but their average is in neither.
$\{(x, z) : x^{2k} \leq z, \, x \geq 0\}$ is convex; this is the set in part 1, intersected with $\mathbb{R}_{\geq 0} \times \mathbb{R}$, which trivially a convex set. The intersection of convex sets is convex.
$\{(x, z) : x^{2k + 1} \leq z, \, x \geq 0 \}$ is convex for all integers $k$. This follows by noting that $\max\{0, x\}^{2k + 1}$ is a convex function and $\mathbb{R}_{\geq 0} \times \mathbb{R}$ is a convex set, and the intersection of convex sets is convex.
and 6. are modifications of 3. and 4., but with strict inequality. By inspection, the strict inequality does not change these arguments (e.g. $\mathbb{R}_{> 0} \times \mathbb{R}$ is still convex.)