Parity of Base and Exponent

It's been established that $y^{2}$ is odd (because adding $665$ , an odd number, to $2^{x}$ results to an odd number), and since odd square numbers are results of odd numbers being squared, $y$ must be odd as well.

To know whether $x$ is even or odd, we need to know the last digits of the sums of $5$ and $2^{x}$ :

$\begin{array}{lc} & \small \text{last digit of sum} \\[0.25em] x=1,5,9,... & \implies 7 \\ x=2,6,10,... & \implies 9 \\ x=3,7,11,... & \implies 3 \\ x=4,8,12,... & \implies 1 \end{array}$

Now let's see the last digits of $y^{2}$ with $y$ being odd:

$\begin{array}{lc} & \small \text{last digit of square} \\ y=1,11,21,... & \implies 1 \\ y=3,13,23,... & \implies 9 \\ y=5,15,25,... & \implies 5 \\ y=7,17,27,... & \implies 9 \\ y=9,19,29,... & \implies 1 \end{array}$

We see that if $x$ is even, the last digits are either $9$ or $1$ which are the same last digits if $y$ ends with the digits $1,3,7,$ and $9$ . Thus, $x$ must be even.

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I accidentally deleted my solution and now I can't post another. So I'm commenting here.

@Chris Lewis, those are the only solutions.