It is also possible to use the (exceptional) outer automorphism of order $2$ of $S_{6}$ to give an "explanation" of why $S_{6}$ is not $\{2,3\}$-generated, along the lines I used in comments for $S_{5}$ above. Take a $6$-cycle $\sigma \in S_{6}.$ Then $\sigma^{2}$ is a product of two disjoint three cycles and $\sigma^{3}$ is a product of three disjoint $2$-cycles. These clearly commute. Now take an outer automorphism $\tau$ of $S_{6}$ which sends products of two disjoint three cycles to three cycles. Then $\tau$ must also send products of three disjoint transpositions to transpositions, since $\tau(\sigma^{2})$ and $\tau(\sigma^{3})$ must commute.

Now suppose that $S_{6}$ is $\{2,3\}$-generated say $S_{6} = \langle \alpha, \beta : \alpha^{2} = \beta^{3} = 1 \rangle.$ Then we may apply $\tau$ if necessary, and assume that $\beta$ is a $3$-cycle. Then $\alpha$ is an odd permutation, so is either a transposition, or a product of three disjoint transpositions.

In the former case, we have a contradiction since there is a point fixed by both $\alpha$ and $\beta$. In the latter case, none of the transpositions in $\alpha$ can fix all points moved by $\beta$, for otherwise that transposition would be central in $\langle \alpha, \beta \rangle.$ It follows that $\alpha$ sends each point moved by $\beta$ to a point fixed by $\beta$ and conversely. It follows that $\beta$ and $\beta^{\alpha}$ commute. Now $\alpha$ normalizes the Abelian subgroup $\langle \beta, \beta^{\alpha}\rangle $, so that $\langle \alpha, \beta \rangle = \langle \alpha \rangle \langle \beta^{\alpha} ,\beta \rangle$ has order dividing $18$, a contradiction.

I do not know if there is an argument using the fact that $S_{8}$ is isomorphic to ${\rm GL}(4,2)\langle \gamma \rangle$, where $\gamma$ is the transpose inverse automorphism, to "explain" that $S_{8}$ is not $\{2,3\}$-generated.

Later edit: It would have been better perhaps to use the outer automorphism of $S_{6}$ to reduce to the case that $\alpha$ is a transposition,(in which case, generation requires that $\beta$ is a product of two disjoint $3$-cycles), and then note the general fact that when $n >1$, $S_{2n}$ is never generated by a transposition $\alpha$ and an element $\beta$ which
is a product of two disjoint $n$-cycles. For if it were, we may conjugate the pair and assume that $\alpha = (12).$ If either of the $n$-cycles in $\beta$ were disjoint from $\alpha$, then that $n$-cycle would be central in $\langle \alpha, \beta \rangle = S_{2n}$, a contradiction. Hence both $n$-cycles of $\beta$ contain a point moved by $\alpha$.

We may conjugate $\beta$ by a permutation fixing both $1$ and $2$ and assume that $\beta = (1357 \ldots 2n-1)(2468 \ldots 2n)$ without disturbing the generation property. Then $\langle \alpha, \alpha^{\beta}, \ldots, \alpha^{\beta^{n-1}}\rangle$ = $\langle (12),(34), \ldots , (2n-1 2n) \rangle$ is Abelian and normal in $\langle \alpha, \beta \rangle = S_{2n},$ a contradiction.

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