Solution I Let $(a, b)$ be a pair of positive integers satisfying the condition. Since $k = \frac{a^2}{2ab^2 - b^3 + 1} > 0$, we have $2ab^2 - b^3 + 1 > 0$ or $a > \frac{b}{2} - \frac{1}{2b^2}$, and hence $a \ge \frac{b}{2}$. Using this, we infer from $k \ge 1$, or $a^2 \ge b^2(2a-b)+1$, that $a^2 > b^2(2a-b) \ge 0$. Hence
$$a > b \text{ or } 2a = b. \qquad \textcircled{1}$$
Now consider the two solutions $a_1, a_2$ of the equation
$$a^2 - 2kb^2a + k(b^3 - 1) = 0 \qquad \textcircled{2}$$
for any fixed positive integers $k$ and $b$, and assume that one of them is an integer. Then the other is also an integer because $a_1 + a_2 = 2kb^2$. We may assume that $a_1 \ge a_2$, and we have $a_1 \ge kb^2 > 0$. Furthermore, since $a_1 a_2 = k(b^3 - 1)$, we get
$$0 \le a_2 = \frac{k(b^3 - 1)}{a_1} \le \frac{k(b^3 - 1)}{kb^2} < b.$$
Together with (1), we conclude that $a_2 = 0$ or $a_2 = \frac{b}{2}$ (in the latter case $b$ must be even).
If $a_2 = 0$, then $b^3 - 1 = 0$, and hence $a_1 = 2k$ and $b = 1$.
If $a_2 = \frac{b}{2}$, then $k = \frac{b^2}{4}$ and $a_1 = \frac{b^4}{2} - \frac{b}{2}$.
Therefore the only possibilities are
$$(a, b) = (2l, 1), (l, 2l) \text{ or } (8l^4 - l, 2l)$$
for some positive integer $l$. All of these pairs satisfy the given condition.

Solution II If $b=1$, it follows from the given condition that $a$ must be even.
Let $\frac{a^2}{2ab^2 - b^3 + 1} = k$. If $b > 1$, then there are two solutions to the equation (②) and one of them is a positive integer. Thus the discriminant $\Delta$ of the equation (②) is a perfect square, that is $\Delta = 4k^2b^4 - 4k(b^3 - 1)$ is a perfect square.
Note that, if $b \ge 2$, we have
$$(2kb^2 - b - 1)^2 < \Delta < (2kb^2 - b + 1)^2. \quad \textcircled{3}$$
The proof is given as follows,
$$\begin{align*} \Delta - (2kb^2 - b - 1)^2 &= 4kb^2 - b^2 - 2b + 4k - 1 \\ &= (4k - 1)(b^2 + 1) - 2b \\ &\ge 2(4k - 1)b - 2b > 0, \end{align*}$$
$$\begin{align*} (2kb^2 - b + 1)^2 - \Delta &= 4kb^2 - 4k - (b-1)^2 \\ &= 4k(b^2 - 1) - (b-1)^2 \\ &> (4k-1)(b^2-1) > 0, \end{align*}$$
this completes the proof of (3).
Since $\Delta$ is a perfect square, it follows from (3) that
$$\Delta = 4k^2b^4 - 4k(b^3 - 1) = (2kb^2 - b)^2.$$
Then $4k = b^2$, and hence $b$ must be even. Let $b = 2l$. We have $k = l^2$. Together with ②, we have $a = l$ or $8l^4 - l$.
Therefore the only possibilities are
$(a, b) = (2l, 1), (l, 2l) \text{ or } (8l^4 - l, 2l)$
for some positive integer $l$. All of these pairs satisfy the given condition.

First Solution. (Based on work by Anders Kaseorg) Rewrite equation $(*)$ as $a^2 - 2ab^2k = -b^3k + k$. Adding $b^4k^2$ to both sides completes the square on the left-hand side and gives
$$(kb^2 - a)^2 = b^4k^2 - b^3k + k,$$
or
$$(2kb^2 - 2a)^2 = (2b^2k)^2 - 2b(2b^2k) + 4k.$$
Completing the square on the right-hand side gives
$$(2kb^2 - 2a)^2 = (2b^2k - b)^2 + 4k - b^2.$$
or,
$$y^2 - x^2 = 4k - b^2,$$
where $x = 2kb^2 - b$ and $y = 2kb^2 - 2a$.
If $4k = b^2$, then either $x = y$ or $x = -y$. In the former case, $b = 2a$; in the latter case, $4kb^2 - b = 2a$, that is, $b^4 - b = 2a$. Because $k = b^2/4$ is an integer if and only if $b$ is even, we get the solutions
$$\left(\frac{b}{2}, b\right) \quad \text{and} \quad \left(\frac{b^4 - b}{2}, b\right)$$
for any even $b$; that is, $(a, b) = (t, 2t)$ and $(a, b) = (8t^4 - t, 2t)$ for all positive integers $t$.
If $4k < b^2$, then $y^2 < x^2$, so $y^2 \le (x-1)^2$ (since $x$ is clearly positive). Thus,
$$4k - b^2 \le (x - 1)^2 - x^2 = -2x + 1 = -4kb^2 + 2b + 1,$$
or, $4k(b^2 + 1) \le b^2 + 2b + 1 < 3b^2 + 1$. Because $4k > 3$, this is a contradiction.
Similarly, if $4k > b^2$, then $y^2 > x^2$, so $y^2 \ge (x+1)^2$. Thus,
$$4k - b^2 \ge (x + 1)^2 - x^2 = 2x + 1 = 4kb^2 - 2b + 1,$$
or $4k(b^2 - 1) + (b - 1)^2 \le 0$. We must have $b = 1$ and
$$k = \frac{a^2}{2a - 1 + 1} = \frac{a}{2}.$$
This is an integer whenever $a$ is even, so we get the solutions $(a, b) = (2t, 1)$ for all positive integers $t$.

Second Solution. (Based on work by Po-Ru Loh) Assume that $b = 1$. Then
$$\frac{a^2}{2ab^2 - b^3 + 1} = \frac{a^2}{2a} = \frac{a}{2}$$
is a positive integer if and only if $a$ is even. Thus, $(a, b) = (2t, 1)$ are solutions of the problem for all positive integers $t$.
Now we assume that $b > 1$. Viewing equation $(*)$ as a quadratic in $a$, replace $a$ by $x$ to consider the equation
$$x^2 - 2b^2kx + (b^3 - 1)k = 0 \quad (*)'$$
for fixed positive integers $b$ and $k$. Its roots are
$$x = \frac{2b^2k \pm \sqrt{4b^4k^2 - 4b^3k + 4k}}{2} = b^2k \pm \sqrt{b^4k^2 - b^3k + k}.$$
Assume that $x_1 = a$ is an integer root of equation $(*')$. Then $b^4k^2 - b^3k + k$ must be a perfect square. We claim that
$$\left(b^2k - \frac{b}{2} - \frac{1}{2}\right)^2 < b^4k^2 - b^3k + k < \left(b^2k - \frac{b}{2} + \frac{1}{2}\right)^2.$$
Note first that
$$\begin{aligned} \left(b^2 k - \frac{b}{2} - \frac{1}{2}\right)^2 &= b^4 k^2 - 2b^2 k \left(\frac{b}{2} + \frac{1}{2}\right)^2 + \frac{1}{4}(b+1)^2 \\ &= b^4 k^2 - b^3 k - b^2 k + \frac{1}{4}(b+1)^2. \end{aligned}$$
To establish the first inequality in our claim, it suffices to show that
$$-b^2 k + \frac{1}{4}(b+1)^2 < k,$$
or, $(b+1)^2 < 4(b^2+1)k$, which is evident as $(b+1)^2 < 2(b^2+1)$ and $k \ge 1$.
Note also that
$$\begin{aligned} \left(b^2 k - \frac{b}{2} + 1\right)^2 &= b^4 k^2 + 2b^2 k \left(\frac{1}{2} - \frac{b}{2}\right) + \frac{1}{4}(1-b)^2 \\ &> b^4 k^2 - b^3 k + b^2 k \\ &> b^4 k^2 - b^3 k + k, \quad \text{as } b > 1, \end{aligned}$$
which establishes the second inequality in our claim.
Because all of $b, k$, and $\sqrt{b^4 k^2 - b^3 k + k}$ are positive integers, we conclude from our claim that $b^2 k - \frac{b}{2}$ is an integer and that
$$b^4 k^2 - b^3 k + k = \left(b^2 k - \frac{b}{2}\right)^2 = b^4 k - b^3 k + \frac{b^2}{4},$$
and so $k = b^2/4$. Thus, $b = 2t$ for some positive integer $t$. The two solutions of the equation $(*)$ becomes
$$x = b^2 k \pm \left(b^2 k - \frac{b}{2}\right),$$
that is. $x = t$ or $x = 8t^4 - t$. Hence, $(a, b) = (t, 2t)$ and $(a, b) = (8t^4 - t, 2t)$ are the possible solutions of the problem, in addition to the solutions $(2t, 1)$.

Third Solution. Because both $k$ and $a^2$ are positive, $2ab^2 - b^3 + 1 > 0$, or,
$$2a > b - \frac{1}{b^2}.$$
Because $a$ and $b$ are positive integers, we have $2a \ge b$. Because $k$ is a positive integer, $a^2 \ge 2ab^2 - b^3 + 1$, or, $a^2 \ge b^2(2a - b) + 1$. Because
$$a^2 > b^2(2a - b) \geq 0,$$
we have
$$a > b \quad \text{or} \quad 2a = b. \qquad (\dagger)$$
We consider again the quadratic equation $(*)$ for fixed positive integers $b$ and $k$, and assume that $x_1 = a$ is an integer root of equation $(*)$. Then the other root $x_2$ is also an integer because $x_1 + x_2 = 2b^2k$. Without loss of generality, we assume that $x_1 \geq x_2$. Then $x_1 \geq b^2k > 0$. Furthermore, because $x_1x_2 = (b^3 - 1)k$, we obtain
$$0 \leq x_2 = \frac{(b^3 - 1)k}{x_1} \leq \frac{(b^3 - 1)k}{b^2k} < b.$$
If $x_2 = 0$, then $b^3 - 1 = 0$, and so $x_1 = 2k$ and $(a, b)$ can be written in the form of $(2t, 1)$ for some integers $t$.
If $x_2 > 0$, then $(a, b) = (x_2, b)$ is a pair of positive integers satisfying the equations $(\dagger)$ and $(*)$. We conclude that $2x_2 = b$, and so
$$k = \frac{x_2^2}{2x_2b^2 - b^3 + 1} = x_2^2 = \frac{b^2}{4},$$
and $x_1 = b^4/2 - b/2$. Thus, $(a, b)$ can be written in the form of either $(t, 2t)$ or $(8t^3 - t, 2t)$ for some positive integers $t$.