The only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $\mathbb{Z}[\sqrt{-2}]$ is $x=5$

$x + \sqrt{-2} = (c + d\sqrt{-2})^3$ leads to $c^3-6cd^2=x$ and $3c^2d-2d^3=1$. The last equation gives $d=\pm1$ and $c=\pm1$. Then the answer is $x=5$.

Directly

$$(a+b\sqrt{-2})^2(a+b\sqrt{-2})=(a^2-2b^2+2ab\sqrt{-2})(a+b\sqrt{-2})=$$

$$=(a^3-6ab^2)+(3a^2b-2b^2)\sqrt{-2}$$

We get that it must be

$$3a^2b-2b^2=1\iff b(3a^2-2b)=1\iff b,\,3a^2-2b=\pm 1$$

But $b=-1\implies 3a^2-2(-1)=-1\implies3a^2=-3$ , impossible, so it must

$$\;b=1\implies3a^2-2=1\iff 3a^2=3\implies a^2=1\iff a=\pm1\;$$

and thus we want $\;\pm1+\sqrt{-2}\;$ , so that for example

$$(1+\sqrt{-2})^3=1+3\sqrt{-2}-6-2\sqrt{-2}=-5+\sqrt{-2}$$

But $\;-5\notin\Bbb N\;$ , so let us try the other option with $\;a=-1\,,\,\,b=1\;$ :

$$(-1+\sqrt{-2})^3=-1+3\sqrt{-2}+6-2\sqrt{-2}=5+\sqrt{-2}$$

and this time $\;x=5\in\Bbb N\;$ .