For the Fermat equation
$x^p$ + $y^p$ = $z^p$
the discriminant of the Frey curve is
$\Delta$ = $16(xyz)^{2p}$
which satisfies Ribet's level lowering requirements for the conductor. This is key to proving FLT.
For the generalized Fermat equation
$x^p$ + $y^q$ = $z^r$
I suppose the discriminant is
$\Delta$ = $16x^{2p}y^{2q}z^{2r}$
Let $n=pqr$.
Then integers can be found that satisfy $ord_l(\Delta)\equiv 0 \pmod n$
In particular, all integers $(a,b,c)$ such that they are pairwise relatively prime and
$x=a^{qr}$, $y=b^{pr}$, $z=c^{pq}$
appear to satisfy $ord_l(\Delta)\equiv 0 \pmod n$. This is an infinite class of triples for the generalized Fermat equation that do not have non-trivial solutions because of FLT. Whenever you can factor the FLT exponent $n$ into three factors, each of those can be used as an exponent in a generalized Fermat equation.
The important question is a more general case. Even though we just ruled out an infinite class of solutions, it only covers $n=pqr$. What about any $(p,q,r)$ where $p,q,r\ge3$?
Edit 2:
I want to rephrase the question for clarity.
For the generalized Fermat equation
$x^p$ + $y^q$ = $z^r$
The discriminant of the Frey curve is
$\Delta$ = $16x^{2p}y^{2q}z^{2r}$
Silverman's version of Ribet's requirement is
$ord_l(\Delta)\equiv 0 \pmod n$
For the discriminant of the Frey curve for the generalized Fermat equation we can write
$\Delta=\alpha \Delta'$ where
$\Delta'=(abc)^{2 \min(p,q,r)}$
Obviously,
$ord_l(\Delta')\equiv 0 \pmod n$ and
$ord_l(\alpha)\equiv 0 \pmod n$ if $l$ is not a factor of $\alpha$, but
what do we do if $l$ is a factor of $\alpha$? Is there any way to reset and go again on $\alpha$ (so to speak)?
References would be Silverman pp.253-255 of Rational Points on Elliptic Curves, and Ribet's original article "On modular representations of Gal(Q bar/Q) arising from modular forms" Invent. math. 100, 431-476 (1990).