Let $3n+1=a^2$ and $4n+1=b^2$. Then we have $(2a)^2 - 3b^2=1$.

Let's consider the solutions to the standard Pell's equation $$u^2-3v^2=1$$

where $u$ is an even number. Its general solution is given by $$u_k + v_k\sqrt{3}=(2+\sqrt{3})^k$$

It is easy to see that $u_k$ is even if and only if $k$ is odd. Therefore the qualified solutions are $$(u_{2k+1}, v_{2k+1})$$

Meanwhile, we also have $$2 u_{2k+1} =(2+\sqrt{3})^{2k+1}+(2-\sqrt{3})^{2k+1}$$

It follows that $$(a, b)=\Big(\frac{1}{2}u_{2k+1}, v_{2k+1}\Big)\quad\text{and}\quad n=\frac{1}{3}(a^2-1)=\frac{1}{12}(u^2_{2k+1}-4)$$

which yields$$48n = (7+4\sqrt{3})^{2k+1} - (7-4\sqrt{3})^{2k+1} - 14$$

or $$n=\frac{7^{2k+1}-7}{24}+2{2 \choose 2k+1}7^{2k-1}+2\cdot 48 {2 \choose 2k+1} 7^{2k-3}+\cdots$$

It is now obvious that $n$ is divisible by 56.