Let, $(18m^2+1)^x+(7m^2-1)^y=(5m)^z \cdots (1)$
Where $m \equiv 3, 27 \pmod{40}$ and $x, y, z$ are odd. Taking equation (1) modulo $40$ we get- $3^x + 22^y \equiv 15^z \pmod {40} $.
In a paper the author claimed (to see the claim, click here and go to page 1994 for detail, in the proof of lemma 8) that equation $(1)$ may have solution for $y=1$ and no solution for $y>1$.
But what I see is that,
Case: modulo 8:
$3^x + 22^y \equiv 15^z \pmod {40} \implies 4+(-2)^y \equiv 0\pmod 8$, so, from the implication, we see that the congruence may have solution only for $y=2$, but we consider odd $y$, so the claim that equation $(1)$ has no solution for $y>1$ is established, but at the same time, $4+(-2)^y \equiv 0\pmod 8$, has no solution for $y=1$ .
Case: modulo 4:
$3^x+22^y \equiv 15^z \pmod{40} \implies -1+0 \equiv 3 \pmod4 $ is true for $y>1$, but has no solution for $y=1$.
From above it appears to me that, the author should have said that the equation $(1)$ has no solution for $y=1$ also, not only $y>1$, in general $y>0$ has no solution for equation $(1)$
Query: Why author is not considering that $y=1$ has no solution in equation$(1)$?
Earlier in the paragraph:
So at this point we have already used the assumption $y>1$, to conclude that $z$ is odd. When $y=1$, it is still possible for $z$ to be even, and we don't arrive at the equation you're citing to begin with.
(In fact, throughout the paper, $(x,y,z) = (1,1,2)$ will keep being a solution, so it's impossible for us to do anything that would eliminate it.)
Once we've shown $z$ is odd, then in the case $m \equiv 3, 27 \pmod{40}$ we get the equation $3^x + 22^y \equiv 15 \pmod{40}$, and you're right that this not only has no solutions when $y>1$, but also has no solutions for any $y \ge 0$.
But observing this wouldn't help us in any way, since we've already assumed $y>1$. (It also wouldn't help us prove the lemma, since we are trying to eliminate the $y>1$ case for as many $m$ as possible, but we're fine with the $y=1$ case sticking around.)