Your suggestion of breaking up into cases is the right one. I would handle the details differently. Suppose for example that $x. We use the fact that $\sqrt[n]{x^n+y^n}=y\sqrt[n]{1+\left(\frac{x}{y}\right)^n}.$
Since $\dfrac{x}{y}<1$, it is clear that $\displaystyle\lim_{n\to\infty} \left(\frac{x}{y}\right)^n=0$.
So if $x, the limit is $y$. Similarly, if $y, the limit is $x$. We need also to take care of the case $y=x$. Then we want to calculate $\displaystyle\lim_{n\to\infty} x\sqrt[n]{2}$. As $n$ gets large, $\sqrt[n]{2}$ approaches $1$.
The answer is therefore always, as you conjectured, $\max(x,y)$.
Comment: We can also handle the problem without breaking into cases. Note that $\max(x,y) <\sqrt[n]{x^n+y^n} \le \sqrt[n]{2}\max(x,y).$ Then we can obtain the desired limit formula directly by "squeezing," since $\displaystyle\lim_{n\to\infty}\sqrt[n]{2}=1$. This argument is more elegant but perhaps less natural than the division procedure that was used in the main post. The intuition there was that if $x, then for large $n$, the term $x^n$ is negligible compared to $y^n$.