Opposite of Fermat's Last Theorem?

Here's a less algebraic way to interpret your question: Look at things in base $a$. Then the equation $a^x+a^y=a^z$ looks like $1 \ldots 0 + 1 \ldots 0 = 1\ldots0$. (where the number of zeros is $x$, $y$, or $z$) This can only work if we're in base $2$ and the addition carries to produce a $1$ in the next digit.

Clearly we must have $x, y<z$. Suppose $x\not=y$; WLOG, suppose $x<y$. Then this can be rewritten as $$a^x(1+a^{y-x})=a^xa^{z-x},$$ with $z-x, y-x>0$. This gives $$1+ a^{y-x}=a^{z-x};$$ but the right hand side is divisible by $a$, while the left hand side is not.

So we must have $x=y$. But then this yields $$2a^x=a^z,$$ which in turn yields $$2=a^{z-x}$$ with $z-x>0$. So there are no nontrivial solutions.

The equation $a^x+a^y=a^z$ has no solutions in positive integers with $a\gt 2$. For suppose the relation holds. Without loss of generality we may assume $x\le y$. Then $$1+a^{y-x}=a^{z-x}.$$ This is only possible if $y=x$, $a=2$, and $z=x+1$.