Diophantus-2

$\displaystyle 3^x - 1 = 2(1 + 3 + 3^2 + ... + 3^{x-2} + 3^{x-1}) = y^3$

$\displaystyle \Rightarrow 2(1 + 3 + 3^2 + ... + 3^{x-2} + 3^{x-1}) = y^3$

$\displaystyle \Rightarrow 1 + 3 + 3^2 + ... + 3^{x-2} + 3^{x-1} = 4 \times k^3$

$\displaystyle \Rightarrow (1 + 3^2 + 3^4 + ... + 3^{x-4} + 3^{x-2}) = k^3$ This is true for $\displaystyle x = 0,1$

$\displaystyle \Rightarrow 10 (1 + 3^4 + 3^8 + ... + 3^{x-8} + 3^{x-4}) = k^3$ This is never true because the number ends with zero, and if we do this more times then, the sum diverges with a higher rate.

Actually there is an even better solution. Since we have $3^x=y^3+1$ , it can be rewritten as $3^x=(y+1)(y^2-y+1)$ and this implies that either $(y+1)|(y^2-y+1) or (y^2-y+1)|(y+1)$ taking each of them respectively as 2 cases, we have that in case 1 $(y+1)|(y+1)^2-3y$ which furthermore implies that $(y+1)|-3y$ which again implies that $(y+1)|-3(y+1)+3$ which again implies that $(y+1)|3$ so there are 2 solutions:(x,y)=(0,0)(or)(2,2). In case 2, since $(y^2-y+1)|(y+1)$ we can conclude that $y^2-y+1<=y+1$ . So again by solving the inequality and substituting in $3^x=(y+1)(y^2-y+1)$ we get again 2 solutions:(x,y)=(0,0)(or)(2,2). Hence the number of solutions is 2.

Looking for solutions with $x,y \in \mathbb{Z}$ ...

We cannot have $x < 0$ , since that would make $y^3$ nonintegral. Hence $x,y \ge 0$ . Since $3^x = y^3 + 1 = (y+1)(y^2-y+1)$ , we deduce that $y = 3^z - 1$ for some integer $0 \le z \le x$ . Then $3^{x-z} \; = \; y^2 - y + 1 \; = \; 3^{2z} - 3^{z+1} + 3$ If $z \ge 1$ then $3^{x-z} = 3^{z+1}\big(3^{z-1} - 1\big) + 3 \equiv 3 \pmod{9}$ , and hence $x-z = 1$ , so that $3^{z+1}\big(3^{z-1} - 1\big) = 0$ , and hence $3^{z-1} = 1$ , so that $z=1$ , and hence $x=y=2$ .

If $z=0$ then $y=0$ and hence $x=0$ .

Thus the only solutions in integers are $(x,y) = (2,2)\,,\,(0,0)$ . There are $\boxed{2}$ solutions in integers.