Here's a proof that the $x=1$ solution is unique
using only facts about "Pell equations" that were already known to Fermat 
(if not centuries earlier to Bhaskara II et al.)
and should generalize at least to $p^x + 2 = y^2$ 
for odd primes $p$ such that $p+2$ is a square. 
I'll simplify the formulas by using $\sqrt 7$ (more generally, $\sqrt p$).

We start by observing that $x$ must be odd; this can be seen
in various ways, for example by reducing $\bmod 4$.  So, $7^x = 7 z^2$
where $z = 7^{(x-1)/2}$, and we seek solutions of $y^2 - 7 z^2 = 2$
where $z$ is a power of $7$.  Now the general solution of $y^2 - 7z^2 = 2$
in natural numbers $y,z$ is $(y_n,z_n)$ such that
$$
y_n + z_n \sqrt7 = \alpha u^n \quad (n \geq 0)
$$
where $\alpha = 3+\sqrt 7$ and $u$ is the fundamental unit
$\alpha / \bar\alpha = 8 + 3\sqrt 7$.  (Note that this gives
$y_n - z_n \sqrt 7 = \alpha u^{-(n+1)}$, so the solutions coming from $n < 0$
only repeat those with $n \geq 0$.)  Now if $x > 1$ then $7 \mid z$,
which happens if and only if $7 \mid 2n+1$.  But then $z_n$ is a multiple of
$z_3 = 7 \cdot 617$, and $617$ is not a multiple of $7$. 
Therefore $z_n$ is never a power of $7$ other than 1, **QED**.

(Alternatively, having found the condition $7 \mid 2n+1$ for $7 \mid z_n$,
we could continue by asking when $7^2 \mid z_n$, $\ 7^3 \mid z_n$,
etc., finding that $7^e | z_n$ if and only if $7^e | 2n+1$, 
and then $z_n \geq z_{(7^e-1)/2}$ which is much larger than $7^e$.)