Task 6.

$$
x^{2}=(y+1)^{2}+1, \quad x^{2}-(y+1)^{2}=1, \quad(x-y-1) \cdot(x+y+1)=1
$$

Since $x, y$ are integers, the equality is possible in two cases:

1) $\left\{\begin{array}{l}x-y-1=1 \\ x+y+1=1\end{array}\right.$
2) $\left\{\begin{array}{l}x-y-1=-1 \\ x+y+1=-1\end{array}\right.$

From the first system, we find that $x=1, y=-1$.

From the second $x=-1, y=-1$.
Answer: $(1,-1),(-1 ;-1)$.