Let $2n+1 = u^2$ and $3n+1=v^2$.

On one hand, $u^2 + v^2 = 5n+2 \equiv 2 \pmod{5}$ implies $u^2 \equiv v^2 \equiv 1 \pmod{5}$. This because a sqaure cannot congruent to $2$ modulo $5$, otherwise it must end with either $2$ or $7$. This means $n$ is a multiple of $5$.

On the other hand, we have $3u^2 - 2v^2 = 1$. Setting $u=x+2y$ and $v=x+3y$, the equation becomes $x^2 - 6y^2=1$. It has infinitely many positive integer solutions.

Since $3u^2-2v^2=1$, $u$ must be odd. Let it be $u=2k+1$. Then $n=2k^2 + 2k$ is even. Since $3n+1=v^2$, $v$ is odd too. Let it be $v=4m\pm 1$. Then $3n=16m^2\pm 8m$, which implies $n$ is a multiple of $8$.

Thus $n$ must be a multiple of 40.