I guess by "skew ring" you mean "division ring". Let $A^*=A \setminus \{0\}.$ Suppose first that $A$ is a division ring. Then, since $A$ is a finite ring, $A$ is a finite field, by the Wedderburn's little theorem. Let $|A|=q.$ Then $x^{q-1}=1$ for all $x \in A^*$ and so $x^{q-1}+y^{q-1}=z^{q-1}$ has no solution in $A^*.$ Conversely, suppose that $A$ is not a division ring (equivalently, a field because $A$ is finite). So $|A| > 2$ and hence the equation $x+y=z$ has solutions in $A^*$ (just choose $y=1, \ z \ne 0,1$ and $x=z-1$). Let $J(A)$ be the Jacobson radical of $A.$ Since $A$ is finite, it is Artinian and so $J(A)$ is nilpotent. So if $J(A) \neq (0),$ then there exists $a \in A^*$ such that $a^2=0$ and so $a^n=0$ for all $n \ge 2.$ Thus $x^n+y^n=z^n, \ n \ge 2,$ has a solution $x=a, y=z=1.$ If $J(A)=(0),$ then by the Artin-Wedderburn's theorem, $$A=\prod_{i=1}^k M_{n_i}(F_i),$$for some finite fields $F_i.$ Since $A$ is not a field, we have either $n_i >1$ for some $i$ or $n_i=1$ for all $i$ and $k \ge 2.$ If $n_i > 1$ for some $i,$ then $M_{n_i}(F_i),$ and hence $A,$ will have a non-zero nilpotent element and we are done. If $n_i=1$ for all $i$ and $k \ge 2,$ then $$x=(1,0,0, \cdots ,0), \ y = (0,1,0, \cdots , 0), \ z = (1,1,0, \cdots , 0)$$will satisfy $x^n+y^n=z^n.$

I will assume: 1. Rings have unity. 2. "Nontrivial" means $x,\,y,\,z\neq 0$. Let the ring be $R$. ($\Longleftarrow$) Suppose FTSOC $R$ is a division ring. Then we can take $n:=|R|-1=|R^\times|$, so $a^n=1$ for all $a\in R\setminus\{0\}$ by Lagrange's theorem. Since $1+1\neq 1$, $x^n+y^n=z^n$ has no solutions, a contradiction. ($\Longrightarrow$) $n=1$ is trivial because $|R|\geq 3$. So assume $n\geq 2$. Since $R$ is not a division ring, there exists a non-unit $a\in R$. Then $1\not\in aR$ so $aR$ has cardinality at most $|R|-1$. Hence there exists $b_1,\,b_2\in R$ with $b_1\neq b_2$ for which $ab_1=ab_2$. Let $b:=b_2-b_1\neq 0$ so $ab=0$. If $b=-a$, we have $a^n+a^n=0=a^n$. If $ba\neq 0$, then $(ba)^2=0$ so we have $(ba)^n+(ba)^n=0=(ba)^n$. If $ba=0$ and $b\neq -a$, we have $(a+b)^n=a^n+b^n$. $\square$