All the possible positive integers $x,y$ such that $x^3 - y^3$ is a perfect square and $x^2 - y^2$ is a perfect cube can be described in terms of reducing $x/y$ to lowest terms $r/s$.
That is, suppose $x = cr$ and $y = cs$ with $r,s$ coprime. Since $x \gt y$, we choose $r > s$.
First we can uniquely factor $r^3 - s^3 = z^2 t$ where $t$ is squarefree.
Furthermore we can uniquely factor $r^2 - s^2 = u^3 v^2 w$ where $v,w$ are squarefree and coprime.
We show that $c$ is uniquely determined up to an arbitrary sixth power factor. That is:
$c = (c_1 c_2^2 c_3^3 c_4^4 c_5^5)*d^6$
where the factors $c_i$ are squarefree, pairwise coprime, and to be determined below, and $d$ is any positive integer.
Substituting for $x,y$ in $x^3 - y^3$ and $x^2 - y^2$, we get these expressions:
$ (cr)^3 - (cs)^3 = (c_1^3 c_2^6 c_3^9 c_4^{12} c_5^{15}) d^{18} (z^2 t) $ $ (cr)^2 - (cs)^2 = (c_1^2 c_2^4 c_3^6 c_4^{8} c_5^{10}) d^{12} (u^3 v^2 w) $
Removing evident squares from the first right-hand side, resp. cubes from the second, we get these conditions that must be satisfied:
$(i) \;\; c_1 c_3 c_5 t$ is a perfect square
$(ii) \;\; c_1^2 c_2 c_4^2 c_5 v^2 w$ is a perfect cube
Using coprimality and squarefreeness of the various factors, the following expressions for the factors $c_i$ are implied:
$c_1 = \gcd(t,w)$
$c_5 = \gcd(t,v)$
$c_3 = t/(c_1 c_5)$
$c_2 = v/c_5$
$c_4 = w/c_1$
Proof:
We claim that $t = c_1 c_3 c_5$ follows from $c_1 c_3 c_5 t$ being a perfect square, in light of the squarefreeness of each factor and the pairwise coprimality of the $c_i$. For any prime factor of $t$ appears to a first power there (owing to the squarefreeness of $t$) and must appear an odd number of times in $c_1 c_3 c_5$ in order for the product of that with $t$ to be a perfect square. But any prime factor of $c_1 c_3 c_5$ can only occur once because the $c_i$ are pairwise coprime and squarefree. Thus the prime factors of $t$ are distinct and in correspondence with the (distinct) prime factors of $c_1 c_3 c_5$. Their equality follows.
By similar arguments we deduce from $c_1^2 c_2 c_4^2 c_5 v^2 w$ being a perfect cube, in light of the squarefreeness of $v,w$ as well as the $c_i$ and coprimality of $v,w$ (as well as the pairwise coprimality of the $c_i$), that $v = c_2 c_5$ and $w = c_1 c_4$.
Now clearly $c_1 = \gcd(t,w)$ and $c_5 = \gcd(t,v)$. The rest of the expressions, for $c_2,c_3,c_4$, are then forced. QED
Examples
For any coprime pair $r \gt s$ we can find a smallest solution of the form $x = cr$ and $y = cs$, and all other solutions where $x/y = r/s$ come from scaling up by an arbitrary sixth power.
The pairs ScottT cites in the Question are among those for which $(r,s) = (10,3)$, and those in a Comment underneath for which $(r,s) = (74,47)$.
To illustrate let's first pick a simple coprime pair, say $(r,s) = (2,1)$.
Then $r^3 - s^3 = z^2 t$ with $t$ squarefree means $t = 7$ and $z = 1$.
Similarly $r^2 - s^2 = u^3 v^2 w$ with $v,w$ squarefree and coprime means $u = 1$, $v = 1$, $w = 3$.
The smallest solution is given by $c = c_1 c_2^2 c_3^3 c_4^4 c_5^5$ where:
$c_1 = \gcd(t,w) = 1$
$c_5 = \gcd(t,v) = 1$
$c_3 = t/(c_1 c_5) = 7$
$c_2 = v/c_5 = 1$
$c_4 = w/c_1 = 3$
Thus the smallest $c = 7^3 * 3^4 = 27783$, which corresponds to $x = 55566$ and $y = 27783$. Verifying:
$x^3 - y^3 = (r^3 - s^3)c^3 = 7 * (7^3 * 3^4)^3 = 7^{10} * 3^{12}$
which is a perfect square, and:
$x^2 - y^2 = (r^2 - s^2)c^2 = 3 * (7^3 * 3^4)^2 = 7^6 * 3^9$
which is a perfect cube. All other solutions with $x/y = 2/1$ are obtained from this one by scaling with a factor $d^6$.
To reprise ScottT's first example $(r,s) = (5,3)$ gives $t = 2$, $v = 1$, and $w = 2$. Then the smallest $c = 2$ gives $x = 10$ and $y = 6$, for which $x^3 - y^3 = 28^2$ and $x^2 - y^2 = 4^3$.
ScottT's second example $(r,s) = (74,47)$ gives:
$74^3 - 47^3 = 301401 = 549^2$
so $t = 1$, and:
$74^2 - 47^2 = 3267 = 3^3 * 11^2$
so $v = 11$ and $w = 1$. The smallest $c = 11^2$ gives $x = 8954$ and $y = 5687$. Note $x^3 - y^3 = 730719^2$ and $x^2 - y^2 = 363^3$.