22. Let $a$ and $b$ be positive integers for which $\frac{a^{2}}{2 a b^{2}-b^{3}+1}=k$ is a positive integer. Since $k>0$, it follows that $2 a b^{2} \geq b^{3}$, so $2 a \geq b$. If $2 a > b$, then from $2 a b^{2}-b^{3}+1>0$ we see that $a^{2}>b^{2}(2 a-b)+1>b^{2}$, i.e., $a > b$. Therefore, if $a \leq b$, then $a = b / 2$. We can rewrite the given equation as a quadratic equation in $a$, $a^{2} - 2 k b^{2} a + k(b^{3} - 1) = 0$, which has two solutions, say $a_{1}$ and $a_{2}$, one of which is in $\mathbb{N}_{0}$. From $a_{1} + a_{2} = 2 k b^{2}$ and $a_{1} a_{2} = k(b^{3} - 1)$ it follows that the other solution is also in $\mathbb{N}_{0}$. Suppose w.l.o.g. that $a_{1} \geq a_{2}$. Then $a_{1} \geq k b^{2}$ and
$$0 \leq a_{2} = \frac{k(b^{3} - 1)}{a_{1}} \leq \frac{k(b^{3} - 1)}{k b^{2}} < b.$$
By the above considerations we have either $a_{2} = 0$ or $a_{2} = b / 2$. If $a_{2} = 0$, then $b^{3} - 1 = 0$ and hence $a_{1} = 2 k, b = 1$. If $a_{2} = b / 2$, then $b = 2 t$ for some $t$, and $k = b^{2} / 4, a_{1} = b^{4} / 2 - b / 2$. Therefore the only solutions are
$$(a, b) \in \left\{(2 t, 1), (t, 2 t), \left(8 t^{4} - t, 2 t\right) \mid t \in \mathbb{N}\right\}$$
It is easy to show that all of these pairs satisfy the given condition.