Question: I was working on an analysis problem when this question arose in one my proofs. I think it may be either $-1$ or $1$, but it seems like there can only be an arbitrary way to assign this.

So is there an agreed upon method for determining $(-1)^x$ for irrational values of $x$?

Answer: 
The general idea is that $a^b$ is to be defined as $\exp(b\log a)$, where that’s the natural logarithm in the parentheses. The problem is that the log has no unambiguous definition for numbers that are not positive real; generally there’s an ambiguity of multiples of $2\pi i$. Since “the” log of $-1$ is $(2k+1)\pi i$ for all integers $k$, you get infinitely many values for $(-1)^x$ when $x$ is not a rational number.
You could think in this:
$$(-1)^x=e^{i\pi x}$$

After a brief commentary actually one really have is:
$$-1=e^{i(\pi +2k\pi)}, k \in \mathbb{Z}$$
And then:
$$(-1)^x=e^{i(\pi +2k\pi)x}, k \in \mathbb{Z} $$
And one would have more than one value for $(-1)^x$