I think we can completely solve the problem. Since $\displaystyle\frac{3n+3x+n^2}{12}$ must be an integer, $y:=3x$ must be an integer. Put also $N:=n(n+3)$. Then we want
$$\frac{N+y}{12},\text{ and }\frac{N(N+y)}{12y}$$ to be positive integers.
Therefore we need $N+y=12r$ (1) and $y=\displaystyle\frac{N^2}{12k-N}$ (2), with $r,k$ integers.
Now, we have $n=4m+1$, so $N=4(m+1)(4m+1)$, which is already a multiple of $4$. Due to (1), $y$ must be also a multiple of $4$. Moreover, if:
- $m\equiv 2 (\text{mod} 3)$ then $N\equiv 0(\text{mod} 12)$ and we need $y\equiv 0(\text{mod} 12)$, so $y\equiv 0(\text{mod} 3)$.
- $m\equiv 1 (\text{mod} 3)$ then $N\equiv 1(\text{mod} 12)$ and we need $y\equiv 2(\text{mod} 12)$, so $y\equiv 2(\text{mod} 3)$.
- $m\equiv 0 (\text{mod} 3)$ then $N\equiv 1(\text{mod} 12)$ and we need $y\equiv 2(\text{mod} 12)$, so $y\equiv 2(\text{mod} 3)$.
Let us find a valid value of $y$ from (2). The first two cases are a bit simpler. We have:
$$y=4\frac{(m+1)^2(4m+1)^2}{3k-(m+1)(4m+1)}.$$
- $m\equiv 2 (\text{mod} 3)$: $m+1,4m+1$ are multiples of 3, hence for $y$ to be a multiple of 3 it is enough to leave one of the factors in the numerator (we choose the smallest one, $m+1$): pick $3k=(m+1)(4m+1)+(m+1)(4m+1)^2$ in order to get $y=4(m+1)$. This can be done since $(m+1)(4m+1)+(m+1)(4m+1)^2$ is a multiple of $3$.
We get $y=4m+4=n+3$ for $n=12t+9$.
- $m\equiv 1 (\text{mod} 3)$: $m+1,4m+1$ are congruent to 2 modulo 3, hence for $y$ to be 2 mod3 it is enough to leave one of the factors in the numerator (we choose the smallest one, $m+1$): pick $3k=(m+1)(4m+1)+(m+1)(4m+1)^2$ in order to get $y=4(m+1)$. This can be done since $(m+1)(4m+1)+(m+1)(4m+1)^2$ is a multiple of $3$. Observe that this is not a happy coincidence: the numerator is congruent to 1 mod3, while the denominator is congruent to -1(mod 3)=2(mod 3), so that $y$ is congruent to $1/2=2$, as needed, and we just need to ensure that $k$ is such that the numerator is a multiple of the denominator.
We get $y=4m+4=n+3$ for $n=12t+5$.
- $m\equiv 0 (\text{mod} 3)$: This one is trickier. $m+1,4m+1$ are congruent to 1 modulo 3, hence the fraction (ignoring the factor of 4) is congruent to 1 mod3, and we need $y$ congruent to 2. We can factor out a 2 from the hanging 4 to get the desired congruence, but then we are not sure anymore that $y$ is indeed a multiple of 4. To be so, either $m+1$ or $4m+1$ must be even, and hence it is necessary and sufficient that $m$ be odd. Thus:
If $m$ is odd we pick $3k=(m+1)(4m+1)+2(m+1)(4m+1)^2$ in order to get $y=2(m+1)=2(\frac{n-1}4 +1)$ for $n=24t+13$.
If $m$ is even, i.e., if $n=24t+1$, then there is no solution.