Recently I considered the equation 
$$x^y+y^z=z^x\qquad \ (x,y,z\in\{1,2,3,\ldots\}),\tag{1}$$
and noted that $(x,y,z)=(1,1,2)$ is a trivial solution. 
I searched solutions of $(1)$ with $x,y,z\le 256$ and did not find another solution. So I made the following conjecture.

**Conjecture 1.** The equation $(1)$ has no solution $(x,y,z)\not=(1,1,2)$.

If $(1)$ has a solution with $x,y,z\in\{3,4,\ldots\}$, then by Beal's conjecture we should have $\gcd(x,y,z)\ge 2$. In light of Fermat's Last Theorem, $(1)$ cannot have a solution with $\gcd(x,y,z)\ge3$.

I aslo considered the equation
$$x^yy^z=z^x\qquad(x,y,z\in\{2,3,\ldots\}).\tag{2}$$
The equation $(2)$ has infinitely many solutions with $x=z$ including $$(x,y,z)=(n^n,n^{n-1},n^n),\ (n^{2n^2},n^{2(n^2-1)},n^{2n^2})\quad \ (n=2,3,\ldots).$$
Actually, all the positive solutions of the equation $x^yy^x=x^x$ are given by
$$x=\prod_{i=1}^rp_i^{a_i\prod_{j=1}^r\ p_j^{a_j}}\ \ \mbox{and}\ \ 
y=\prod_{i=1}^rp_i^{a_i(\prod_{j=1}^r \ p_j^{a_j}-1)},$$
where $p_1,\ldots,p_r$ are distinct primes, and $a_1,\ldots,a_r$ are nonnegative integers. This can be easily deduced by using the Fundamental Theorem of Arithmetic. What about solutions of $(2)$ with $x\not=z$? I have made the following conjecture.

**Conjecture 2.** The only solutions of $(2)$ with $x\not=z$ are as follows:
\begin{align}(x,y,z)=&(8,2,2),\, (9,3,3),\, (4,2,8),\, (16,4,32),\, \\&(128,64,32),\,(512,256,32),\, (686,392,98).\end{align}

**QUESTION.** How to solve Conjectures 1 and 2?

Your are welcome to make comments or check Conjectures 1 and 2 further.