When is $32x+32$ a square number?

$$32(x + 1) = y^2 \implies 32 \mid y^2 \implies 4 \mid y$$ so let $y = 4z$. $$2(x+1) = z^2 \implies 2 \mid z$$ so let $z = 2w$ $$x + 1 = 2w^2 $$ so given any $w$ we can solve for $x$.

So the set of solutions is $$\{(2w^2 - 1, 8w) \mid w \in \Bbb Z\}$$

The answer by user3491648 gives a complete characterization of the full set of values $x$ for which $32x+32$ is a square numbers. This is really just an observation on the set of values

$$\{64,1600,53824,1827904,62094400,2109381184\}$$

reported by the OP.

These are not actually values that $x$ takes on. (As user3491648 finds, it requires on odd value of $x$ to make $32x+32$ a square.) Rather, these seem to be the square values that correspond to values of $x$ that are themselves squares, namely of the following values:

$$\{1,7,41,239,1393,8119\}$$

For example, $32\cdot239^2+32=1827904$. It's unclear why these values are being singled out, but they do seem to obey a two-term recursion, namely

$$a_{n+1}=6a_n-a_{n-1}$$

For example, $1393=6\cdot239-41$.