Set $a+b+2=x,b+1=y$ and we have that $\dfrac{x+1}{y}+\dfrac{x+1}{y+1}=1+\dfrac{6}{x-1}, y\neq 0,-1,x\neq 1$ $....\Leftrightarrow (x-1)(x+1)(2y+1)=3y(y+1)(x+1)$ so $x=-1$ or $x\neq -1$ and $(x-1)(2y+1)=3y(y+1)\Rightarrow 2y+1|3y^2+3y|6y^2+6y\Rightarrow $ $\Rightarrow 2y+1|( 6y^2+6y)-(6y^2+3y)=3y\Rightarrow...\Rightarrow 2y+1|3$ Since $y\neq 0,-1$ we get that $y=1,-2$ and thus $(x-1)\cdot 3=3\cdot 2\Leftrightarrow x=3$ or $(x-1)\cdot(-3)=3\cdot (-2)(-1)=6\Leftrightarrow x-1=-2\Leftrightarrow x=-1,contradiction$ In conclusion, we have the solutions $(a,b)=(-3-t,t),t\in \mathbb{Z}- \left \{0,-1 \right \}$ and $(a,b)=(1,0)$