Thank you! More generally, can we use a similar argument if we replace the cubic power by a larger power, say (1+x^2)^t-1? Would we have a growing number of cases to verify manually? The cases t = (p-1)/2 and t = p-1 are special and can be resolved directly. When $t \ne (p-1)/2, p-1$, I always find a lot of solutions for any p.

@NoamD.Elkies Yes, the finite field of $p$ elements.

I edited the question to clarify.

Sorry. For the equivalent problem, I am assuming -1 is not a square in $\mathbb{Z}_p$, i.e., $p \equiv 3 \pmod 4$. That's the case I am the most interested in, but I believe the first statement holds for all $p > 3$.