One more solution of the Mordell equation

[Update: Improved second code.]

There is a system limit on Solve, which you can extend this way:

k = 1000000;
n = Ceiling[k^(3/2)];
With[{ropts = SystemOptions["ReduceOptions"]},
  Internal`WithLocalSettings[
   SetSystemOptions[
    "ReduceOptions" -> "SolveDiscreteSolutionBound" -> n],
   Solve[x^3 - y^2 == 307 && -k < x < k && 0 < y < n, {x, y}, 
    Integers],
   SetSystemOptions[ropts]
   ]] // AbsoluteTiming
(*
  {143.664,
    {{x -> 7, y -> 6}, {x -> 11, y -> 32},
     {x -> 71, y -> 598}, {x -> 939787, y -> 911054064}}}
*)

For speed using an exhaustive search over x: The code will work efficiently for machine integers (for solutions with x^3 less than 2^53, the limit on double-precision floating-point numbers to exactly represent an integer).

Block[{Part},
   With[{x = #[[1]] + 1, y = #[[2]]},
    Hold[
     Pick[#[[All, 1 ;; 2]], #[[All, -1]], 0] &@
      NestList[
       With[{n = Sqrt[x^3 - 307.]},
         If[FractionalPart@n == 0,
          {x, Round[n], 0},
          {x, y, 1}]
         ] &,
       {Floor@CubeRoot@307., 1, 1},
       1000000
       ]
     ]
    ]
   ] // ReleaseHold // AbsoluteTiming
(*  {0.36922, {{7, 6}, {11, 32}, {71, 598}, {939787, 911054064}}}  *)

If you want a brute-force approach to check the rectangular {x, y} space, keep in mind that for 0 <= x <= 10^6, the space has 10^15 pairs, which would take a long time for a GHz processor, or even a few thousand of them.

I used this code to find one. In every iteration I looked up 200k range.

m = 100000;
Total@Boole[IntegerQ /@ Sqrt[Range[8 m, 10 m]^3 - 307]]

1

And extracted the solution using

Position[IntegerQ /@ Sqrt[Range[8 m, 10 m]^3 - 307], True]

{{139788}}

This implies that

x=139788 - 1 + 8 m=939787

is a solution.

{x,y}={939787,911054064}

Since I used Brute-Force, next solution must be $x>10,000,000$