Suppose $n^2 + 24n + 35$ is a square. Let $n^2 + 24n + 35 = m^2$. This equation can be written as $(2n + 2m + 24)(2n - 2m + 24) = 436$. Since $436 = 2^2 \times 109$, we have $\{2n + 2m + 24, 2n - 2m + 24\} = \{1, 436\}, \{-1, -436\}, \{2, 218\}, \{-2, -218\}, \{4, 109\}, \{-4, -109\}$.

For instance, we have $2n + 2m + 24 = 1$, $2n - 2m + 24 = 436$ giving $n = 389/4$, $m = -435/4$, and $2n + 2m + 24 = 436$, $2n - 2m + 24 = 1$ giving $n = 389/4$, $m = 435/4$. Since the solutions are not integers, they are rejected.

Solving $2n + 2m + 24 = 2$, $2n - 2m + 24 = 218$ gives $n = 43$, $m = -54$, and solving $2n + 2m + 24 = -2$, $2n - 2m + 24 = -218$ gives $n = -67$, $m = 54$. The other systems do not give integer solutions.

Thus the only solutions are $n = 43$ and $-67$. In both cases, the square is $54^2 = 2916$.