Let $I_m=[m!+2,m!+m]\cap\mathbb N$ an "interval" of $\mathbb N$; it can obviously be as long as we want and it is easy to prove $I_m$ does not contain any prime. Prove the following: $$\text { if }n^2+(n+1)^2\in I_m\text{ then }4n^2+1\notin I_M$$
Note that if in a large interval $I_m$ could exist $n$ denying what is proposed here, then we would have found a counterexample to the conjecture in here
On
Assume that $2n^2+2n+1 \in [x!+2,x!+x]$ and we want to show that $4n^2+1 \not \in [y!+2,y!+y]$ for some $y>x$.
First see that $4n^2+1 <2(2n^2+2n+1)$ so $\frac{4n^2+1}{2n^2+2n+1} \leq 2$.
Secondly see that $y!+2 > x! +x $ for all $x>2$ , because at least $y=x+1$ so $(x+1) * x! +2 > 2 x! +2x $ for all $x>2$ thus $\frac{y!+2}{x!+x} > 2$.
Which means that $\frac{4n^2+1}{2n^2+2n+1} < 2 < \frac{y!+2}{x!+x}$ in other words if $2n^2+2n+1 \in [x!+2,x!+x]$ then $4n^2+1 \not\in [y!+2,y!+y]$.
Remark(I): At first notice that: $\dfrac{3+\sqrt{9+4}}{2} \leq \dfrac{4+4}{2}=4$ , so for $4 \leq n$ we have:
$$0 \leq n^2-3n+1 \ \ \Longrightarrow \ \ 3n^2+3n+2 \leq 4n^2+1 \ \ \Longrightarrow \ \ \\ \dfrac{3}{2}\Big(2n^2+2n+1 \Big) < 4n^2+1 \ \ \Longrightarrow \ \ \dfrac{3}{2}\Big(n^2+(n+1)^2 \Big) < 4n^2+1 . $$
Remark(II): On the other hand let $4 \leq m$, then we have:
$2 < (m-1)!$, i.e. $1 < \dfrac{(m-1)!}{2}$ multiplying both sides by $m$ we get:
$$m < \dfrac{m!}{2} \ \ \Longrightarrow \ \ m!+m < m!+ \dfrac{m!}{2} = \dfrac{3}{2} m! \ \ \ \ . $$
Suppose on contrary that $4n^2+1 \in I_m $.
Now notice that sicce both of $n^2+(n+1)^2$ and $4n^2+1$
belongs to the interval $[m!+2,m!+m]$, therefor we have:
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ m! < m!+2 \leq n^2+(n+1)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(III)} \ , \ \ \ \ \ \ \ \ \ \ \ \text {and} \ \ \ \ \\ 4n^2+1 \leq m!+m \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(IV)} .$$
$\color{Red}{\text{First case}}$: Let $m$ and $n$ are both greater or equal than $4$,
i.e. $4 \leq m$ and $4 \leq n$.
In this case we have:
$$\dfrac{3}{2} m! \overset{ \tiny{ \text{III} } }{<} \dfrac{3}{2}\Big(n^2+(n+1)^2 \Big) \overset{ \tiny{ \text{Rmk(I)} } }{<} 4n^2+1 \overset{ \tiny{ \text{IV} } }{\leq} m!+m \overset{ \tiny{ \text{Rmk(II)} } }{<} \dfrac{3}{2} m! \ \ \ \ , $$
so we have: $\dfrac{3}{2} m! < \dfrac{3}{2} m!$ ,which is an obvious contradiction. So this case is immpossible!
$\color{Red}{\text{Second case}}$: Let $4 \leq m$ and $n \leq 3$.
In this case by the (III) inequality we have:
$$26= 4!+2 \leq m! + 2 \overset{ \tiny{ \text{III} } }{\leq} \Big(n^2+(n+1)^2 \Big) \leq 9+16=25 \ , $$
which is again an obvious contradiction!
$\color{Red}{\text{Third case}}$: Let $m \leq 3$ and $4 \leq n$.
In this case by the (IV) inequality we have:
$$265= 4.(4)^2+1 \leq 4n^2+1 \overset{ \tiny{ \text{IV} } }{\leq} m!+m \leq 3!+3=9 \ ,$$
which is again an obvious contradiction!
$\color{Red}{\text{Fourth case}}$: Let $m \leq 3$ and $n \leq 3$.
In this case we have the following sub-cases:
$m=3$, then we have: $I_3=[8,9]=\{ 8, 9 \}$. So $n^2+(n+1)^2=8$ or $n^2+(n+1)^2=9$, but none of them have a solution.
$m=2$, then we have: $I_2=[6,6]=\{ 6 \}$. So $n^2+(n+1)^2=6$ , but it does'nt have a solution.
$m=1$, then we have: $I_1=[3,2]=\phi$.
At the end, it looks, that it was better; if I have been organized the cases as follows:
$\color{Green}{\text{First case}}$: $\color{Yellow}{4 \leq m}$ and $4 \leq n$.
$\color{Green}{\text{Second case}}$: $\color{Yellow}{4 \leq m}$ and $n \leq 3$.
$\color{Purple}{\text{Third case}}$: $\color{Yellow}{m} \color{Orange}{\leq} \color{Yellow}{3}$