Answer: $(0,1)$ and $(-1,0)$.

Since $a$ and $a+1$ are coprime, $a^2$ and $a+1$ are also coprime. Similarly $b^2$ and $b-1$ are coprime. Hence the equality can hold only in the case $a+1 = \pm b^2$ and $b-1 = \pm a^2$, where the signs in both equations are the same.

Let both signs be pluses. Then from the first equation we get $a = b^2 - 1 = (b-1)(b+1)$. The second equation implies $b-1 = a^2$, whence $a = a^2(a^2+2)$. If $a=0$, then $b=1$, i.e. $(a,b) = (0,1)$. If $a \neq 0$, then by dividing by $a$ we get $1 = a(a^2+2)$; since $a^2+2 > 1$, this equation does not have integer solutions.

If both signs are minuses then by multiplying by $-1$ we get $-b+1 = a^2$ and $-a-1 = b^2$. These are the same equations with respect to $-b$ and $-a$ which we had previously with respect to $a$ and $b$, hence the only solution is $-b = 0, -a = 1$ i.e. $(a,b) = (-1,0)$.