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An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

The source for Frits Beukers's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here [a credible but perhaps not quite an "official source"].

An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

The source for Frits Beukers's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here.

An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

The source for Frits Beuker's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here [a credible but perhaps not quite an "official source"].

An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

The source for Frits Beukers's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here [a credible but perhaps not quite an "official source"].

An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".

The source for Frits Beuker's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here [a credible but perhaps not quite an "official source"].