Inspired by the ones you found we can see that there are infinitely many solutions as follows: $$(x,y,z;n) = (k-1,\quad k(k-1),\quad k-1;\quad 2^k)$$ for any $k\ge 0$.

Question 1: Inspired by the ones you found we can see that there are infinitely many solutions as follows: $$(x,y,z;n) = (k-1,\quad k(k-1),\quad k-1;\quad 2^k)$$ for any $k\ge 0$.

Edit re: Question 2: How about instead of $$n^{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}+n^{\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}}+n^{\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_1}}=n^{\frac{1}{x_4}+\frac{1}{x_1}+\frac{1}{x_2}}$$ you make it $$n^{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}+n^{\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}}=n^{\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_1}}+n^{\frac{1}{x_4}+\frac{1}{x_1}+\frac{1}{x_2}},$$ so that it has greater symmetry.

Inspired by the ones you found we can see that there are infinitely many solutions as follows: $$(x,y,z;n) = (k-1,\quad k(k-1),\quad k-1;\quad 2^k)$$ for any $k\ge 0$.

Namely, by plugging in this reduces to $$2\cdot 2^{k(1/(k-1)+1/(k(k-1))} = 2^{k(1/(k-1)+1/(k-1))},$$ $$1+k(1/(k-1)+1/(k(k-1)) = k(1/(k-1)+1/(k-1)),$$ $$k-1+k+1=k+k.$$

Inspired by the ones you found we can see that there are infinitely many solutions as follows: $$(x,y,z;n) = (k-1,\quad k(k-1),\quad k-1;\quad 2^k)$$ for any $k\ge 0$.