Let $(m+1)^3-m^3=n^2$, then $(2n)^2 - 3(2m+1)^2 = 1$, so $(2n, 2m+1)$ is a solution to $x^2-3y^2=1$.

Hence, there exists a positive integer $p$ such that $2n+(2m+1)\sqrt{3}=(2+\sqrt{3})^p$.

Now in order for $n$ to be an integer, $p$ must be odd. Let $p=2k+1$ where $k$ is a positive integer. It follows that $$4n=(2+\sqrt{3})^{2k+1} + (2-\sqrt{3})^{2k+1}$$ Finally, $$2n-1 = \frac{(1+\sqrt{3})^2(2+\sqrt{3})^{2k}+(1-\sqrt{3})^2(2-\sqrt{3})^{2k}-8}{2}=N^2$$

where $N=\frac{1}{2}\big((1+\sqrt{3})(2+\sqrt{3})^k+(1-\sqrt{3})(2-\sqrt{3})^k\big)$ is an integer.