Zero and One are the only Consecutive Perfect Squares

Theorem
If $$n$$ is a perfect square other than $$0$$, then $$n+1$$ is not a perfect square.

Proof
Let $$x$$ and $$h$$ be integers such that $$x^2 + 1 = (x - h)^2$$

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Consecutive Integers are Coprime, but both sides must have the same unique prime factorization by the Fundamental Theorem of Arithmetic, so $$h$$ cannot have any prime factors since they cannot be shared by $$(h - 1)(h + 1)$$.

This leaves $$h = -1$$, $$h = 0$$, or $$h = 1$$ as the only possibilities since they are the only integers with no prime factors.

If $$h = -1$$ then $$h + 1 = 0$$, so $$2xh = 0$$. It follows that $$x = 0$$.

If $$h = 1$$ then $$h - 1 = 0$$, so $$2xh = 0$$. It follows that $$x = 0$$.

If $$h = 0$$, then $$2x\cdot 0 = (-1)(1)$$, a contradiction.

Therefore the only pairs of consecutive perfect squares are $$0^2 = 0$$ and $$(0 + (-1))^2 = (-1)^2 = 1$$, and $$0^2 = 0$$ and $$(0 + 1)^2 = 1^2 = 1$$.