Triangular Number cannot be Cube

Theorem
Let $T_n$ be the $n$th triangular number such that $n > 1$.

Then $T_n$ cannot be a cube.

Proof
Suppose $T_n = x^3$ for some $x \in \Z$.

Then by Odd Square is Eight Triangles Plus One:
 * $\exists y \in \Z: 8 T_n + 1 = \paren {2 x}^3 + 1 = y^2$

By Cube which is One Less than a Square:
 * $2 x = 2$, $y = 3$

giving the unique solution:
 * $T_n = 1^3 = 1$