By the symmetry of the problem we may suppose $a\le b$. Notice that $b^2-1\ge 0$ so that if $\dfrac{a^2+b}{b^2-a}$ is a positive integer, then $a^2+b\ge b^2-a$. Rearranging this inequality and factoring we find that $(a+b)(a-b+1)\ge 0$. Since $a,\,b>0$, we must have $a\ge b-1$. We therefore have two cases:
Case 1: $a=b$. Substituting, we have
$\dfrac{a^2+a}{a^2-a}=\dfrac{a+1}{a-1}=1+\dfrac{2}{a-1}$, which is an integer iff $(a-1)|2$. As $a>0$, the only possible values are $a-1=1 \text{or} 2$. Hence, $(a,\,b)=(2,\,2)$ or $(3,\,3)$.
Case 2: $a=b-1$. Substituting, we have
$\dfrac{b^2+a}{a^2-b}=\dfrac{(a+1)^2+a}{a^2-(a+1)}=\dfrac{a^2+3a+1}{a^2-a-1}=1+\dfrac{4a+2}{a^2-a-1}$
Once again, notice that $4a+2>0$ and hence for $\dfrac{4a+2}{a^2-a-1}$ to be an integer, we must have $4a+2\ge a^2-a-1$, that is $a^2-5a-3\le 0$. Hence, since $a$ is an integer, we can bound $a$ by $1\le a \le 5$. Checking all ordered pairs, we find that only $(a,\,b)=(1,\,2)$ or $(2,\,3)$ satisfy the given conditions.
Thus, the ordered pairs that work are $(a,\,b)=(2,\,2),\,(3,\,3),\,(1,\,2),\,(2,\,3),\,(2,\,1),\,(3,\,2)$
