Here
https://en.wikipedia.org/wiki/Fermat%E2%80%93Catalan_conjecture
the fermat catalan-conjecture is mentioned. In the german version, I read that in the case $$\frac{1}{m}+\frac{1}{n}+\frac{1}{k}=1$$ satisfied by the triples $(2/4/4)$ , $(2/3/6)$ , $(3/3/3)$ , the equation $$a^m+b^n=c^k$$ with coprime positive integers $a,b,c$ has finite many solutions. In the cases $(2/4/4)$ and $(3/3/3)$ no solution exists.
What about the cae $(2/3/6)$ ? Which solutions do exist in this case ?
This answer is heavily based on ABCs of Number theory by Noam Elkies. You can read the whole article at https://dash.harvard.edu/bitstream/handle/1/2793857/Elkies%20-%20ABCs%20of%20Number%20Theory.pdf?sequence=2 .
Your case of $(2, 3, 6)$ is equivalent to the rational points of Elliptic curve $Y^2=X^3 \pm 1$. They have rank $0$ (https://en.wikipedia.org/wiki/Rank_of_an_elliptic_curve), so there are finitely many rational points. One can check that the only nonzero rational point is $(X,Y)=(2,3)$, and the solutions of your equation set is trivial multiples of $2^3+1^6=3^2$.