Find all such $b\in\mathbb{Z}_+$ that for all $n\in\mathbb{Z}_+$ there exists $y\in\mathbb{Z}_+$ such that $b^n+1=2y^3$

The question is how to find all such positive integer $b\in\mathbb{Z}_+$ that for any positive integer $n\in\mathbb{Z}_+$ there exists a positive integer $y\in\mathbb{Z}_+$ such that the following equality stands: $$b^n+1=2y^3$$

Well, I tried $b=1$ and it fits. Also, I figured that $b$ must be odd :)

• It never works for $b>1$. You only have to consider the case $n=3$ and use a result of Euler that there is no solution with $b>1$ to $b^3+1=2y^3.$ – user940 Oct 27 '16 at 20:55