Integer solutions of $x^y-z^3=2$

Question

Is it an open question to solve $x^y-z^3=2$ in integers (both positive, zero and negative)? If not, what kind of methods the solution requires?

Answer 1

If $y$ is even, the only solutions to $-t^3 + u^2 = 2$ are $(-1,\pm 1)$

E_+00002: r = 1   t = 1   #III =  1
          E(Q) = <(-1, 1)>
          R =   0.7545769032
           2 integral points
            1. (-1, 1) = 1 * (-1, 1)
            2. (-1, -1) = -(-1, 1)

http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+

Things do not change very much if you allow $-2,$ all you get is $27 - 25 = 2.$

E_-00002: r = 1   t = 1   #III =  1
          E(Q) = <(3, 5)>
          R =   1.3495768357
           2 integral points
            1. (3, 5) = 1 * (3, 5)
            2. (3, -5) = -(3, 5)

http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL-

Also, $y$ cannot be divisible by $3,$ as you can factor the difference of cubes and get a finite number of things to rule out. So, you have $y=5,7,11,13,\ldots$

My guess is that there is a book somewhere on Catalan's conjecture, now proved, with a chapter on changing the difference between powers from $\pm 1$ to $\pm 2, \pm 3,$ and a few other small numbers.

Whatever the exact contents, there is a book called Catalan's Conjecture by Rene Schoof, http://oskicat.berkeley.edu/record=b16478355~S1