Solve $y = x^N + z^N$ for N? Is there a way to solve equations of the form 
$y = x^N+z^N$ for N?
I've been looking and I don't think there is a way, but I'm not sure if there's some obscure algebra rule that will help me here.

As already said, there is no closed form for the solution (except for very specific cases) and only numerical methods would solve the problem.
Let us admit that $x,y,z$ are positive numbers and $N$ be a real number and consider Newton method which, starting from a reasonable guess $N_0$, will update it according to $$N_{k+1}=N_k-\frac{f(N_k)}{f'(N_k)}$$ So, if $$f(N)=x^N+z^N-y$$ $$f'(N)=x^N \log (x)+z^N \log (z)$$ Concerning $N_0$, you could look at the graph to get a rough estimate or, even simpler, start in the middle of the interval suggested by marty cohen.
For illustration purposes, let us use $x=2,z=5,y=100$. Using the bounds suggested by marty cohen, we find $$\frac{\log (50)}{\log (5)}<N<\frac{\log (50)}{\log (2)}$$ So, let us start at the middle point, that is to say $N_0=4$ and start iterating. The following iterates will then be generated : $3.46804$, $3.06212$, $2.85830$, $2.81737$, $2.81598$ which is the solution for six significant figures.
We can do better (obtaining a much more linear function) if instead we search for the root of $$g(N)=\log(x^N+z^N)-\log(y)$$ $$g'(N)=\frac{x^N \log (x)+z^N \log (z)}{x^N+z^N}$$ Starting again with $N_0=4$, the iterates are : $2.82901$, $2.81598$.
Function $g(N)$ is so close to linearity that computing its value at the bounds and assuming that a straight line joins the points gives a very good estimate of the solution. In the case used above for illustration purposes, this would give $N\approx 2.80472$ which is indead very close to the solution. So, if $N$ is supposed to be an integer, this could be a very fast method.
Doing the same using now $y=\frac{1}{100}$, the iterations would start at $N_0=-5.5$ and the iterates are $-6.64344$, $-6.64712$ which is the solution for six significant figures. Note that the estimate given by the above described secant would provide an estimate equal to $-6.66046$.
Edit
By the way, you could apply the same method for solving $$\sum_{i=1}^m a_i^N=y$$ for any number of terms. Admitting $a_1<a_2<\cdots<a_m$, The solution will be such that $$\frac{y}{m \log (a_m)}<N<\frac{y}{m \log (a_1)}$$ Let us try using $m=4,a_1=2,a_2=3,a_3=5,a_4=7,y=6000$. The secant gives an estimate which is $\approx 4.66956$ from where we can start Newton iterations for the function $$g(N)=\log\Big(\sum_{i=1}^m a_i^N\Big)-\log(y)$$ and obtain as iterates : $4.35247$, $4.35166$ which is the solution for six significant figures.
