2024/043) Given $x^2 = 2^ y + 2023$ find x+ y where x and y are natual numbers(IIT JEE 2024)

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 $x+y=46$