Consider $(m+3)^4$ if $2|(m+3)^4$, then $4(m+3)^4|n(n+1)(n+2)(n+3)$. If $3|(m+3)^4$, then $3(m+3)^4|n(n+1)(n+2)(n+3)$, For all cases $(m+3)^4|n+i$ for some $0\le i\le 3$. Then $n(n+1)(n+2)(n+3)=(a(m+3)^4-i)(a(m+3)^4-i+1)(a(m+3)^4-i+2)(a(m+3)^4+3-i)>(m+3)^4((m+3)^4-1)((m+3)^4-2)((m+3)^4-3)>m(m+1)^2(m+2)^3(m+3)^4.$

Rust wrote: Consider $(m+3)^4$ if $2|(m+3)^4$, then $4(m+3)^4|n(n+1)(n+2)(n+3)$. If $3|(m+3)^4$, then $3(m+3)^4|n(n+1)(n+2)(n+3)$, For all cases $(m+3)^4|n+i$ for some $0\le i\le 3$. Then $n(n+1)(n+2)(n+3)=(a(m+3)^4-i)(a(m+3)^4-i+1)(a(m+3)^4-i+2)(a(m+3)^4+3-i)>(m+3)^4((m+3)^4-1)((m+3)^4-2)((m+3)^4-3)>m(m+1)^2(m+2)^3(m+3)^4.$ can you write in more details ?

I am sorry, my proof work only for case $m+3=p^l$, when $m+3$ had only one prime divisor. I think better, if $x=n^2+3n+1$ and $x^2=1+m(m+1)^2(m+2)^3(m+3)^4=m^{10}+20m^9+...=f(m)^2+g(m)$, were $deg(f(m))=5,deg(g(m))\le 4$. Therefore $f(m)^2+g(m)<(f(m)+1)^2$ for $m>m_0$. Then we can chek $m\le m_9$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=379570&p=2099851#p2099851

The equation is equivalent to: $$(n^2 + 3n + 1)^2 = m(m+2) [(m+1)(m+2)(m+3)^2]^2 + 1.$$ If we set $a = n^2 + 3n+1$ and $b = (m+1)(m+2)(m+3)^2$, then $(a, b)$ is a solution of the Pell equation $a^2 = m(m+2) b^2 + 1.$ This has fundamental equation $(m+1, 1)$, and so we know that $(a + b \sqrt{m(m+2)}) = (m+1 + \sqrt{m(m+2)})^k$ for some $k \in \mathbb{N}.$ By size considerations, it's easy to eliminate the cases where $k \ge 5,$ because $n^2 + 3n + 1 \le (m+3)^{\frac52}.$ For $k \le 4$, we must only check small $m$, which is easy to do. $\square$