This is an alternative proof of the aforementioned fact.
Let $2n+1=p^2$ and $3n+1=q^2$. Then, $$ n=q^2-p^2 \tag{1} $$ and $$ 3p^2-2q^2=1\, . \tag{2}$$ Modulo $5$, $(2)$ reduces to $$ 3(p^2+q^2) \equiv 1 \pmod{5}\qquad\hbox{or}\qquad p^2+q^2 \equiv 2 \pmod{5}\, ,$$ which is possible if and only if $p^2 \equiv q^2 \equiv 1 \pmod{5}$.
Modulo $8$, $(2)$ reduces to $2(q^2-1) \equiv 0 \pmod{8}$, which implies that $q$ is odd, and hence $p^2 \equiv q^2 \equiv 1 \pmod{8}$. Thus, $p^2-q^2$ is divisible by both $5$ and $8$, and hence also by $40$.
The proof looks right to me, but I'd like to check. Please help.