Found this a little tricky (despite not taking too long), as one might dive into confusing cases, so I hope this works. If $x=y=z$, then $72x = 2x^5$, so $x=y=z = \sqrt{6}$. From now on suppose that they are not all equal. Summing both sets gives $x^5 + y^5 + z^5 = 36(x+y+z)$. The Power Mean inequality implies $36(x+y+z) > \frac{(x+y+z)^5}{81}$, so $x+y+z < 3\sqrt{6}$. Without loss of generality assume $x$ is smallest among the three, so $x < \sqrt{6}$. Now consider $y^5 + z^5 = ax + by + cz$, where $\{a,b,c\} = \{23, 24, 25\}$. Since $x$ is smallest, the right-hand side does not exceed $36(y+z)$ (e.g. if $a=23$, $b=24$, $c=25$, then we use $12x \leq 12y$ and $11x \leq 11z$). But now $x^5 + y^5 + z^5 = 36(x+y+z)$ implies $x^5 \geq 36x$ and hence $x \geq \sqrt{6}$, contradiction. Therefore there are no other solutions.

I found this easier than p2(maybe because p2 was a combi) By power mean we have $3\sqrt{6}>\sum x $ and then considering the cases where the sets are the same we get that $\sum x^5 \ge 36\sum x$ which is impossible because from the first equation we get $min(x,y,z)<\sqrt{6}$ but from the last one we get $min(x,y,z)\ge \sqrt{6}$ hence the only solution is $x=y=z=\sqrt{6}$. Note:This is a well known trick for problems like this(when the sets are same).If i find a similar question i will edit the post.