We are given
$x^2 = 2^ y + 2023$
Now working mod 4 we have
$2^ y + 2023 \equiv 2^y + 3 \pmod 4$
y cannot be greater than 1 as $2^y + 3 \equiv 3 \pmod 4$
As a square cannot be $3 \pmod 4$ so only possible value is y = 1 giving x= 45 and $x+y=46$
We are given
$x^2 = 2^ y + 2023$
Now working mod 4 we have
$2^ y + 2023 \equiv 2^y + 3 \pmod 4$
y cannot be greater than 1 as $2^y + 3 \equiv 3 \pmod 4$
As a square cannot be $3 \pmod 4$ so only possible value is y = 1 giving x= 45 and $x+y=46$
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