The empty set, $\varnothing$. Trivially, $x=0,1$ yield constant arithmetic progressions; we show that there are no other possibilities. If these numbers do form a progression, then, by the AM-GM (arithmetic mean-geometric mean) inequality, $$2 \cdot 2^{x^{2}}=2^{x}+2^{x^{3}} \geq 2 \sqrt{2^{x} \cdot 2^{x^{3}}} \Rightarrow 2^{x^{2}} \geq 2^{\left(x+x^{3}\right) / 2} \Rightarrow x^{2} \geq\left(x+x^{3}\right) / 2 \Rightarrow x(x-1)^{2}=x^{3}-2 x^{2}+x \leq 0$$ Assuming $x \neq 0,1$, we can divide by $(x-1)^{2}>0$ and obtain $x<0$. However, then $2^{x}, 2^{x^{3}}$ are less than 1, while $2^{x^{2}}$ is more than 1, so the given sequence cannot possibly be an arithmetic progression.