Solution:
Let $d = \gcd(a, b)$ 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 $(a, b) \in \{(2, 12); (4, 6); (6, 4); (12, 2)\}$.