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
(This proof was completed with an insightful comment from Gottfried. I'm placing it as an answer so that it is readily seen that an answer exists, as opposed to just leaving it in the question.)
Suppose we have $ 2^n \pm 1 = x^k$ for some positive integers $x, k$. Cases $n=1, 2, 3$ are easily dealt with. $n=3$ yields the solution $2^3 + 1 = 3^2$. Henceforth, let $n\geq4$. Furthermore, a simple check shows that $x\neq 1, 2$.
First, we will deal with the power $k=2l$ being even. Rewriting $x^{2l}$ as $(x^l)^2$, we may assume that $k=2$.
Proof: $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2 that differ by 2. Thus we must have $(x-1) = 2, (x+1) = 4$, which gives $2^n = 8$ so $n=3$ (reject). $_\square$
Now, we deal with $k$ odd.
Proof: Say $2^n +1 = x^k$. Then $2^n = x^k - 1 = (x-1)(x^{k-1}+x^{k-2} + \ldots +1)$, so both terms are powers of 2. We have $ x -1$ is a power of 2 greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $k$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^k$. Then $2^n = x^k + 1 = (x+1)(x^{k-1} - x^{k-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x+1$ is a power of 2 that is greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction. $ _\square$