Let $d=\operatorname{gcd}(a, b)$ and $x, y \in \mathbb{Z}_{+}$ such that $a=d x, b=d y$. Obviously, $(x, y)=1$. The equation is equivalent to $d^{4} x^{2} y^{2}+208=4 d^{2}(x y+1)^{2}$. Hence $d^{2} \mid 208$ or $d^{2} \mid 13 \cdot 4^{2}$, so $d \in\{1,2,4\}$. Take $t=x y$ with $t \in \mathbb{Z}_{+}$.

Case I. If $d=1$, then $(x y)^{2}+208=4(x y+1)^{2}$ or $3 t^{2}+8 t-204=0$, without solutions.

Case II. If $d=2$, then $16 x^{2} y^{2}+208=16(x y+1)^{2}$ or $t^{2}+13=t^{2}+2 t+1 \Rightarrow t=6$, so $(x, y) \in\{(1,6) ;(2,3) ;(3,2) ;(6,1)\} \Rightarrow(a, b) \in\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\}$.

Case III. If $d=4$, then $16^{2} x^{2} y^{2}+208=4 \cdot 16(x y+1)^{2}$ or $16 t^{2}+13=4(t+1)^{2}$ and if $t \in \mathbb{Z}$, then 13 must be even, contradiction!

Finally, the solutions are $(a, b) \in\{(2,12) ;(4,6) ;(6,4) ;(12 ; 2)\}$.