On a conjecture that if $2\left(p_n+p_{n-1}\right)\equiv -1\pmod {n^2}$ then $n=11$ only. 

Conjecture:

We denote by $p_n$ the $n^{\text{th}}$ prime number. If $2\left(p_n+p_{n-1}\right)\equiv -1\pmod {n^2}$, then $n=11$ only.

How must one be able to prove / disprove this supposed claim? I had come up with this conjecture, but am a little stuck in trying to prove it.

Attempt:
I know that the congruence can be similarly written as, $2\left(p_n+p_{n-1}\right) = mn^2 - 1$ for some $m\in\mathbb{Z}$, the set of all integers. However, I will consider the case where $m=1$, or where $2\left(p_n+p_{n-1}\right)$ $ = (n+1)(n-1)$. Of course now, we find that $n$ must be odd and greater than $1$.
Replace $n$ with $n_k$, namely, the $k^\text{th}$ odd number. Then, $$\frac{n_k+1}{2} = k \ \land \ \frac{n_k-1}{2} = k-1.\tag{$\land$ is read as $and$}$$ Therefore, $p_{n_k}+p_{n_k-1} = k(n_k-1) = (k-1)(n_k+1)$. This means that $\left\{k,k-1\right\}\mid p_{n_k} + p_{n_k-1}$. So, $p_{n_k}+p_{n_k-1}$ has to be squarefree, leaving out primes $3$ and $5$ since $5+3=8=2^3$ and also $17$ and $19$ since $19+17=36=6^2$.
I do not know where to go from here. Am I taking the correct approach? Is there anything important I should know abut prime numbers (e.g. any lemmas and such) that will be useful in (dis)proving my conjecture? Hints are mostly appreciated, but you can provide a full and disclosed answer if you wish.

Thank you in advance.
 A: Not a complete answer due to my lack of explicit bounds, but shows the set of such numbers is finite.
In order for $2\left(p_n + p_{n-1}\right) \equiv -1 \pmod {n^2}$, it is necessary that there exists a $k \in \mathbb{Z}$ such that $2\left(p_n + p_{n-1}\right) = kn^2 - 1$.
In particular this means that $$\frac{2\left(p_n + p_{n-1}\right)+1}{n^2}=k$$ is an integer.
On the other hand, from the Prime Number Theorem we have the estimate $p_n \sim n\log n$, which means $$\frac{2\left(p_n + p_{n-1}\right)+1}{n^2} \sim\frac{4n\log n}{n^2} = \frac{4 \log n}{n}$$
This means that the sequence converges to $0$ and so there is some finite $N$ for which the sequence remains strictly between $0$ and $1$ for all $n \geq N$. Thus there can be only finitely many cases where the desired equality holds.
A Wolfram Alpha plot plot supports the conjecture.
EDIT:
A more explicit upper bound from Wikipedia gives the explicit upper bound valid for $n \geq 6$: $$p_n \lt n(\log n + \log \log n)$$
Using this result we find that $$\frac{2\left(p_n + p_{n-1}\right)+1}{n^2} \lt \frac{5p_n}{n^2} \lt \frac{5n(\log n + \log \log n)}{n^2} = 5\frac{\log n + \log \log n}{n} \lt 1$$ for $n \geq 21$ so it is sufficient to verify the conjecture for all $2 \leq n \leq 20$.
