At first glance I thought that for $(1)$ to have solutions $n$ had to be a multiple of $3$, which is wrong  when $x$ can be a rational, as Gerhard pointed out. So what follows is a solution when $n=3m$ for some $m$. Remain to be treated the cases $n\equiv1 \pmod 3$ and $n\equiv2 \pmod 3$.

So $n=3m$, with $m\equiv3 \pmod 4$.

Writing $m=4p+3$ condition $(1)$ becomes
$$\dfrac{3(4p+3)+x+3(4p+3)^2}{4}\in \mathbb{N}$$

which is equivalent to $$\dfrac{1+x+3}{4}\in \mathbb{N}$$
so that $$(1)\Longleftrightarrow x\in 4\mathbb{N}$$.

Now if you plug in $n=3(4p+3)$ and $x=4q$ into $(2)$, you get massive simplifications into

$$\dfrac{(4p+3)(p+1)(3q+9(4p+3)(p+1))}{q}\in \mathbb{N}$$

showing that for instance any factor of $(4p+3)(p+1)$ will do.

> Conclusion: when $n\equiv0 \pmod 3$
>
> - $p=\dfrac{n-9}{12}$ is an integer, 
> - $x=4(p+1)$ is a solution.