The sum of the first n squares is:  

n(n+1)(2n+1) / 6

If you start with the k-th square and sum the next 2023 of them, the result is

(k+2023)(k+2024)(2k+4047) / 6  -  k(k+1)(2k+1) / 6

((2k^3 + 12141k^2 + 24567313k + 16570651944) - (2k^3 + 3k^2 + k)) / 6

(12138k^2 + 24567312k + 16570651944)/6

(2023k^2 + 4094552k + 2761775324)

2023(k^2 + 2024k + 1365188)

2023(k^2 + 2024k + 1024144  + 341044)

2023((k + 1012)^2 + 341044)

Given 2023(k^2 + 2024k + 1365188), the quadratic must be divisible by 2023 for the expression to be a square.

The last term, 1365188, is 1686 mod 2023

The second term, 2024k mod 2023 is (k mod 2023)

So, (k^2 + k + 1686) must be 0 mod 2023

A brute force program shows that this last expression can take on 548 values, but zero is not one of them.  Thus no value of k satisfies this condition.

The sum of 2023 consecutive squares is never a perfect square.