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: 
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 $(-1)^x$
The usual agreed-apon method, is to imagine that the imaginary part of the exponent is a line wrapped around the circle, the circumference is $2\pi$. The agreed on position is some point either in the range $-\pi\lt x \le\pi$ or $0 \le x \lt 2\pi$.

So if one has a number like $i$, then its complex logrithm is $\pi/2$. If we mean to raise it to some power, like $\pi^{17}$, then one calculates first $(\pi^{17}.\pi/2) \mod 2\pi$, that is, the remainder when an integer multiple of $2\pi$ is subtracted from the calculated value. And this remainder gives the power of the number raised to that power.

The solution for equations like $x^7=1$ is effected by $e^{2\pi ni}x^7=1$ for $n=0$ to $6$, and using the same model as above. When $n$ is not rational, this can go on forever, progressively splattering the unit circle with answers.