$y^2 = x^3 - 26$, exist ideal satisfying conditions? For the solution $(x, y) = (3, 1)$ of $y^2 = x^3 - 26$, 

does there necessarily exist an ideal $I$ of the integer ring $\mathbb{Z}[\sqrt{-26}]$ of $\mathbb{Q}(\sqrt{-26})$ such that $(y + \sqrt{-26}) = (1 + \sqrt{-26})$ is equal to $I^3$ but this $I$ is not principal?


Let $K=\mathbb{Q}(\sqrt{-26})$, and $\sigma$ its non-trivial automorphism. We have ${\cal O}_K=\mathbb{Z}[\sqrt{-26}].$
If $I^3 = (1+\sqrt{-26})$, then 
$$N_{K/\mathbb{Q}}(I)^3=(I I^\sigma)^3 = I^3 (I^\sigma)^3 = (1+\sqrt{-26})(1-\sqrt{-26})=(27)=(3)^3.$$
Then $I I^\sigma=N_{K/\mathbb{Q}}(I)=(3)$. Since $3$ is prime, $I$ (and $I^\sigma$) must be prime ideals.
So you're asking whether 3 splits in $K$, and if so, whether the prime ideal dividing $3$ is non-principal. The answer is yes, and
$$I = (3,1+\sqrt{-26}).$$
It's not hard to see that $II^\sigma \subset (3)$, and $(3)\subset II^\sigma$ follows from
$$ 3 = 3^2-3(1+\sqrt{-26})-3(1-\sqrt{-26}).$$
Then $3=II^\sigma$ and $I$, $I^\sigma$ are prime ideals. Also $I\neq I^\sigma$ since $3$ does not divide $26$.
The prime decomposition of $(1+\sqrt{-26})(1-\sqrt{-26})=(27)$ is $I^3 (I^\sigma)^3$, so the possible prime factors of $(1+\sqrt{-26})$ are $I$ and $I^\sigma$. Since $II^\sigma = (3)$ and $(3)$ does not divide $(1+\sqrt{-26})$, either $(1+\sqrt{-26})=I^3$ or $(I^\sigma)^3$. Since $1+\sqrt{-26}\in I$, $I$ already divides $(1+\sqrt{-26})$, so $(1+\sqrt{-26})=I^3$.
