Let $p=2$ be the even prime and without loss of generality assume $p<q<r$. Since $p,q,r$ are primes this means $r\geq 5$.
We shall first show the following:
Lemma. Let $2=p<q<r$ be primes and
$$
\begin{align}
2q-6 &= x^2\\
2r-6 &= y^2\\
qr-6 &= z^2
\end{align}
$$
for some integers $x,y,z\geq 0$. Then
$$
z+y = r
$$
Proof. Taking equations $2$ and $3$, we form the following:
$$
\begin{align}
z^2 &\equiv -6 \equiv y^2 \pmod r\\
(z+y)(z-y) &\equiv 0 \pmod r
\end{align}
$$
Therefore $r$ divides $z+y$ or $z-y$. We first assume the latter, then
$$
\begin{align}
z-y &\equiv 0 \pmod r\\
z &= y + kr
\end{align}
$$
for some $k\geq 0$ since $z> y$. But now
$$
\begin{align}
0 &< y+kr=z =\sqrt{qr-6} < \sqrt{r^2} = r\\
0 & < y+kr < r \implies k=0
\end{align}
$$
This means that
$$
z = y
$$
which is not possible. Hence we must have instead
$$
\begin{align}
z+y &\equiv 0 \pmod r\\
z+y &= kr\\
0 < kr &= z+y\\
&= \sqrt{qr-6}+\sqrt{2r-6}\\
&< \sqrt{r^2} + \sqrt{4r}\\
&= r+2\sqrt{r}\\
&< 2r\\
0<kr &< 2r
\end{align}
$$
where the last equality is because
$$
r\geq 5 \implies 4r < r^2 \implies 2\sqrt{r} < r
$$
Therefore we must have precisely $k=1$, giving $z+y=r$.
$$
\tag*{$\square$}
$$
We now show that
Proposition. Let $p=2$ and
$$
\begin{align}
2r-6 &= y^2\\
qr-6 &= z^2\\
z+y &= r
\end{align}
$$
for some integers $q,r,y,z\geq 0$. Then
$$
p+q+r-9 = (y-1)^2
$$
Proof. Using $z+y=r$, we have
$$
\begin{align}
r &= z+y\\
&= \sqrt{qr-6} + \sqrt{2r-6}\\
\sqrt{qr-6} &= r - \sqrt{2r-6}\\
qr-6 &= r^2-2r\sqrt{2r-6} + (2r-6)\\
qr &= r^2-2r\sqrt{2r-6} + 2r\\
q &= r -2\sqrt{2r-6} + 2\\
p+q+r-9 = q+r-7 &= (2r-5) -2\sqrt{2r-6}\\
&= (y^2+1) - 2y\\
&= (y-1)^2
\end{align}
$$
$$
\tag*{$\square$}
$$
Edit 1: Deriving a formula similar to Will Jagy's.
Proposition. Primes $q,r$ satisfies
$$
\begin{align}
q &= 2(6k\pm 1)^2+3 = 5 + 24u\\
r &= 2(6k\pm 2)^2+3 = 11 + 24v
\end{align}
$$
for some integers $u,v\geq 0$ (must be same sign). This in turn gives
$$
\begin{align}
x &= \sqrt{2q-6} = 2(6k \pm 1)\\
y &= \sqrt{2r-6} = 2(6k \pm 2)\\
z &= \sqrt{qr-6} = 72k^2\pm 36k+7\\
q+r-9 &= (3(4k\pm 1))^2
\end{align}
$$
Proof. We start from
$$
\begin{align}
2q-6&=x^2\implies -6\equiv x^2\pmod q\\
2r-6&=y^2\implies -6\equiv y^2\pmod r
\end{align}
$$
Since $-6$ is a square, by Quadratic Reciprocity we get $q,r\equiv 1,5,7,11\pmod{24}$.
Now the third equation gives
$$
qr-6 = z^2\implies qr=z^2+6
$$
Since $z^2+6\equiv 6,7,10,15,18,22\pmod{24}$, the only possible combinations of $(q,r)\pmod{24}$ are
$$
(1,7),(5,11),(7,1),(11,5)
$$
For $q\equiv 1,7 \pmod{24}$, we observe that
$$
2q-6 =x^2\implies 20,8\equiv x^2\pmod{24}
$$
which is not possible. Therefore up to interchanging $q$ and $r$, we may assume that
$$
(q,r)\equiv (5,11) \pmod{24}
$$
Hence we now have
$$
\begin{align}
q &= 2a^2+3 = 5+24u\\
r &= 2b^2+3 = 11+24v
\end{align}
$$
for some integers $a,b,u,v\geq 0$. By checking $\pmod{24}$, we can show that
$$
a = 6m\pm 1,\quad b = 6n\pm 2
$$
Further, using $q=r-2\sqrt{2r-6}+2$ as before:
$$
\begin{align}
q &= r-2\sqrt{2r-6}+2\\
72 m^2 \pm 24 m + 5 &= 72 n^2 \pm 24 n + 5\\
9 m^2 \pm 6 m + 1 &= 9 n^2 \pm 6 n + 1\\
(3m \pm 1)^2 &= (3n\pm 1)^2
\end{align}
$$
it must be the case that $m=n=k$ and $6m\pm 1,6n\pm 2$ has the same sign. Then a direct computation gives the formula for $z$ and $p+q+r-9$.
Best Answer
We know that the sum up to $2k-1$ is $k^2$, so the sum up to $313$ is $157^2$. If the highest number in your sum is $2n-1$ we then are asking that $$n^2-157^2=x^2\\(n+x)(n-x)=157^2$$ There will be a solution for each way of factoring $157^2$ into two numbers of the same parity, so both must be odd. As $157$ is prime, the only factorization of interest is $1,157^2$ $$n+x=157^2\\n-x=1\\n=\frac 12(1+157^2)=12325\\x=12324$$