I'm going to put these as a part answer and delete my earlier comments, I'm not sure people click on the "show 5 more comments," it took me weeks to notice that option, plus adding many comments did seem to slow the Latex font resolution.
First, Kevin's initial argument showing that $m$ must be odd is readily reworded to show that
$3^m + 5^m neq 0 pmod p$ for primes $p equiv 7, 11, 43, 59 pmod {60},$ as then
$15, (5/3), (3/5)$ are all quadratic residues $pmod p$ but $-1$ is not, so we simply never get
$ (5/3)^m equiv -1 pmod p$ for these primes.
Now adding in the fact that you have proved $m$ really must be odd in a genuine solution, we find that
$3^m + 5^m neq 0 pmod p$ for primes $p equiv 13, 29, 37, 41 pmod {60},$ as then
$15, (5/3), (3/5)$ are all quadratic nonresidues $pmod p$ but $-1$ is a residue, so with $m$ odd we never get
$ (5/3)^m equiv -1 pmod p$ for these primes either.
Put these together with everything else, we get $m equiv 3,5 pmod 8,$ then $m-1$ and $m+1$ cannot be divisible by 3 or 5 or by any prime $p$ with Legendre symbol $( -15 | p) = -1.$
In a similar spirit to Gjergji Zaimi, corollary to these observations got me as far as showing that $m equiv 3, 93 pmod {120},$ not quite optimal here.
It is also true that $m-1$ and $m+1$ cannot be divisible by 17 or 353 even though $-15$ is a quadratic residue here, these being primes $p equiv 1 pmod 4$ with the property that the smallest $m$ solving $ (5/3)^m equiv -1 pmod p$ happens to be even, hence all possible such $m$ by a standard argument emphasizing the property of minimality. One has $p=17, ; m=2$ and $p=353, ; m=4$ and allowing larger $m$ we have $p=17, ; m equiv 2 pmod 4$ and
$p=353, ; m equiv 4 pmod 8.$
Well, the strategy, a little less foolish than it seemed for a while, is to show that both $m-1$ and $m+1$ fail to be divisible by any odd primes in a genuine solution to the original problem, hence both are powers of 2, hence by inequalities $m=3.$ That is the hope anyway. The smallest uncertain prime is 19.
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