The Lemma in the aswer by Bjorn Poonen can be sharpened to give an exact relationship
between $\mathrm{den}(x)$ and $\mathrm{den}(x')$.

**Lemma 1.**
$~$Let $f=f_2+f_1+f_0\in\mathbb{Z}[X]=\mathbb{Z}[X_1,\ldots,X_n]$ ($n\geq 1$),
with each $f_i$

homogeneous of degree $i$.
Let $y,v\in\mathbb{Z}^n$, where $v$ is primitive and $f_2(v)\neq0$,
and define $F=AT^2+BT+C:=f(y+Tv)\in\mathbb{Z}[T]$.
Suppose that the two zeros $t$ and $t'$ of $F$ are rational,
so that the two rational points $x=y+tv$ and $x'=y+t'v$ are zeros of $f$.
If $x=a/b$ and $x'=a'/b'$ are the reduced representations (with $b,b'>0$),
and $x\neq y$ (which is certainly true if $x\notin\mathbb{Z}^n$),
then
\begin{equation*}
b' \,=\, \mathrm{sgn}(A)\cdot
\frac{f_2(x-y)}
{\mathrm{gcd}(A,B,C)\,\mathrm{gcd}(a-by)^2}\cdot b~; \tag{1}
\end{equation*}
exchanging $x$ and $x'$ gives the analogous identity (provided $y\neq x'$).

**Remark.**
$~$For any $u=(u_1,\ldots,u_n)\in\mathbb{Z}^n$ we write $\mathrm{gcd}(u):=\mathrm{gcd}(u_1,\ldots,u_n)$.

**Proof.** $~$Since $tv=x-y=(a-by)/b=(c/b)v$,
where $c=\pm\,\mathrm{gcd}(a-by)$ and $\mathrm{gcd}(c,b)=1$,
we have $t=c/b$, and similarly $t'=c'/b'$ with $c'=\pm\,\mathrm{gcd}(a'-b'y)$
and $\mathrm{gcd}(c',b')=1$. The leading coefficient of $F$ is $A=f_2(v)=(b/c)^2f_2(x-y)$
($c\neq 0$ because $a-by=b(x-y)\neq 0$).

By Gauss lemma $F=d(bT-c)(b'T-c')$ with $d=\mathrm{sgn}(A)\cdot\mathrm{gcd}(A,B,C)$.

Then $dbb'=A=(b/c)^2f_2(x-y)$ gives us $b'$ expressed
as in the lemma.$~$
**Done.**

The Lemma in Bjorn Poonen's post is an immediate consequence.

Lemma 1 is about the geometric background of Davenport-Cassels lemma:
it relates the reduced representations of two rational zeros of $f$
that lie on an integral line $L=y+\mathbb{Q}v$
whose direction vector $v$ is an anisotropic vector of the quadratic form $f_2$.
(An integral line is an affine line in $\mathbb{Q}^n$
that contains an integral point and hence infinitely many integral points.)
It is not required that one or the other of the two zeros is non-integral,
the lemma says something interesting even when both zeros are integral.
Also, the two zeros may coincide, in which case the line $L$
is tangent to the quadric $\{f=0\}$.

If we actually write out the other identity mentioned in the lemma
and then compare the two identities,
we obtain the identity (supposing $y\neq x,x'$)
\begin{equation*}
f_2(x-y)f_2(x'-y)
\,=\, \bigl(\mathrm{gcd}(A,B,C)\,\mathrm{gcd}(a-by)\,\mathrm{gcd}(a'-b'y)\bigr)^2~.\tag{2}
\end{equation*}
It is not in the least surprising that $f_2(x-y)f_2(x'-y)$ is a square:
if $q$ is any quadratic form on a vector space $V$ over some field $K$,
and $u\in V$ and $\lambda,\mu\in K$,
then it is trivial that $q(\lambda u)q(\mu u)=\bigl(\lambda\mu q(u)\bigr)^2$.
It may seem slightly surprising that $f_2(x-y)f_2(x'-y)$ is a square of an integer,
but this is
a consequence of $A=dbb'$: in the situation of the lemma
the trivial identity satisfied by a general quadratic form reads
\begin{equation*}
f_2(x-y)f_2(x'-y) \,=\, \Bigl(\frac{c}{b}\,\frac{c'}{b'}f_2(v)\Bigr)^2~,
\end{equation*}
and substituting $f_2(v)=A=dbb'$ yields
\begin{equation*}
f_2(x-y)f_2(x'-y) \,=\, (dcc')^2 \,=\, C^2 \,=\, f(y)^2
\end{equation*}
(this time without the restriction $y\neq x,x'$).
Let $L_{\mathbb{Z}}$ denote the set of all integral points on the line $L$:
$L_{\mathbb{Z}}:=L\cap\mathbb{Z}^n=y+\mathbb{Z}v$.
For any $z\in L_{\mathbb{Z}}$
the identity (2) still holds when $y$ is replaced by $z$,
but we must be careful and write it as
\begin{equation*}
f_2(x-z)f_2(x'-z)
\,=\, \bigl(\mathrm{gcd}(A,B_z,C_z)\,\mathrm{gcd}(a-bz)\,\mathrm{gcd}(a'-b'z)\bigr)^2~,
\end{equation*}
because the coefficients $B_z$ and $C_z$ of $F_z=f(z+Tv)$ depend on $z$.
However,
note that the coefficient $A_z=A=f_2(v)$,
as well as the greatest common divisor of the coefficients of $F_z$
(the content of $F_z$),
$\mathrm{gcd}(A,B_z,C_z)=\left|A\right|/bb'=\left|d\right|$, do *not* depend on $z$.
For $z\in L_{\mathbb{Z}}$ we define $c(z),c'(z)\in\mathbb{Z}$
by $x-z=\bigl(c(z)/b\bigr)v$ and $x'-z=\bigl(c'(z)/b'\bigr)v$.
Since $\mathrm{gcd}(a-bz)=\left|c(z)\right|$ and $\mathrm{gcd}(a'-b'z)=\left|c'(z)\right|$,
we have
\begin{equation*}
f_2(x-z)f_2(x'-z) \,=\, \bigl(d\,c(z)\,c'(z)\bigr)^2 \,=\, C_z^2 \,=\, f(z)^2~,
\qquad\quad z\in L_{\mathbb{Z}}\,.
\end{equation*}
**Remark.** $~$Idiot me!
This is just a very special case of the general power-of-a-point theorem,
which does not rely on specific factorization properties of integers
and is almost trivial to prove:

Let $K$ be a field,
let $f\in K[X] = K[X_1,\ldots,X_n]$ be of degree $m$ (where $m, n\geq 1$),
and denote by $f_m$ the homogeneous component of $f$ of degree $m$.
Let $L$ be an affine line in $K^n$ with a direction vector $v$, where $f_m(v)\neq 0$.
Let $y\in L$, and define $F_y := f(y+Tv)\in K[T]$, a polynomial of degree $m$.
Suppose that $F_y$ has $m$ zeros (counting multiplicities) $t_1$, $\ldots$, $t_m$ in $K$.
Then the points $x_i=y+t_iv\in L$, $1\leq i\leq m$, are zeros of $f$,
the multiset of the $x_i$'s does not depend on the choice of $y\in L$,
and
\begin{equation*}
f_m(y-x_1)f_m(y-x_2)\cdots f_m(y-x_m) \,=\, f(y)^m~.
\end{equation*}

The independence is easy: if $z=y+sv$,
then $F_z$ has the zeros $t_i-s$, whence $z+(t_i-s)v = y+t_iv = x_i$.
The leading coefficient of $F_y$ is $f_m(v)$ and its constant term is $f(y)$.
From $F_y=f_m(v)(T-t_1)\cdots(T-t_m)$ we get $f(y)=(-1)^m t_1\cdots t_m f_m(v)$,
whence $f_m(y-x_1)\cdots f_m(y-x_m) = \bigl((-t_1)\cdots(-t_m)f_m(v)\bigr)^m = f(y)^m$.

We digress.
Let's return to the situation in Lemma 1.

We regard the point $y$ as fixed, serving as an origin of $L_{\mathbb{Z}}$.
Let us determine $c(z)$ for a general point $z=y+kv\in L_{\mathbb{Z}}$, $k\in\mathbb{Z}$:
from
\begin{equation*}
\frac{c(y+kv)}{b}\,v \,=\, x-(y+kv) \,=\, (x-y)-kv \,=\, \frac{c(y)-kb}{b}\,v
\end{equation*}
we see that
\begin{equation*}
c(y+kv) \,=\, c(y) - kb~. \tag{3}
\end{equation*}
For any $z\in L_{\mathbb{Z}}$ we have
\begin{equation*}
f_2(x-z) \,=\, \frac{c(z)^2}{b^2}f_2(v)
\,=\, \frac{c(z)^2}{b^2}\,dbb'
\,=\, \frac{db'}{b}\,c(z)^2
\,=\, \frac{e_0}{b_0}\,c(z)^2~, \tag{4}
\end{equation*}
where $b_0=b/\mathrm{gcd}(b,db')$ and $e_0=db'/\mathrm{gcd}(b,db')$.
Since $\mathrm{gcd}\bigl(b,c(z)\bigr) = 1$, and hence $\mathrm{gcd}\bigl(b_0,c(z)\bigr) = 1$, it follows that
\begin{equation*}
\mathrm{den}\bigl(f_2(x-z)\bigr) \,=\, b_0 \qquad\quad \text{for every $z\in L_{\mathbb{Z}}$}\,.
\end{equation*}
Combining (3) and (4) we obtain
\begin{equation*}
f_2\bigl(x-(y+k)v\bigr) \,=\, \frac{e_0}{b_0}\bigl(c(y)-kb\bigr)^2~, \qquad\quad k\in\mathbb{Z}\,.
\end{equation*}
In the special case $x=x'$, when the line $L$ is a tangent of the quadric $\{f=0\}$,
we have $b=b'$, whence
\begin{equation*}
f_2(x-z) \,=\, d\,c(z)^2~, \qquad\quad z\in L_{\mathbb{Z}}\,,
\end{equation*}
thus $f_2(x-z)$ is an integer for every integral point $z$ in $L$.

The discussion above has demonstrated that the identity (1) and its brethren have uses
unrelated to Davenport-Cassels lemma.
Now we return to applications of (1) to Davenport-Cassels lemma and (a little way) beyond it.
The assumptions $x\in\mathbb{Q}^n\setminus\mathbb{Z}^n$ and $0<\left|f_2(x-y)\right|<1$
imply the premises $f_2(v)\neq 0$ and $x\neq y$ of Lemma 1
and then yield the instantaneous result $b'<b$.
But besides the $f_2(x-y)$ in the numerator on the right hand side of (1)
there are also the factors $\mathrm{gcd}(A,B,C)$ and $\mathrm{gcd}(a-by)^2$ in the denominator
whose product can be greater than $1$ and can help make $b'$ smaller than $b$
even when $\left|f_2(x-y)\right|\geq 1$.
This leads to the idea of walking with Aubry 'on the far side':
when there is no integral point $y$ satisfying $\left|f_2(x-y)\right|\geq 1$
we may still find an integral point $y$
so that we can make a step along the line $L$
from a zero $x=a/b$ of $f$ to a zero $x'=a'/b'$ of $f$ with $b'<b$.
The folowing two examples attest that this idea actually works.

But first, a definition.
A quadratic form $q$ on $\mathbb{Q}^n$ is said to be *Euclidean*
if for every $x\in\mathbb{Q}^n\setminus\mathbb{Z}^n$
there exists $y\in\mathbb{Z}^n$ such that $0 < \left|q(x-y)\right| < 1$.

For the first example let $q(X_1,X_2,X_3)=X_1^2+X_2^2+5X_3^2$
and $f(X)=q(X) - m$, where $m$ is a positive integer.
The positive-definite quadratic form $q$ is not Euclidean on $\mathbb{Q}^3$,
since for any $x\in\mathbb{Z}^3+\bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigr)$
and any $y\in\mathbb{Z}^3$ we have $q(x-y)\geq 7/4$.
Consider the set $F$ of all points $x$ in the cube
$\bigl\{(x_1,x_2,x_3)\in\mathbb{R}^3 \bigm| 0\leq x_1,x_2,x_3\leq \frac{1}{2}\bigr\}$
at which $q(x)\geq 1$;
the set $F$ ('the far side') is shown as the darker shaded part of the cube
in the following figure:

The function $q(1-x_1,1-x_2,1-x_3)$ of a point $(x_1,x_2,x_3)$ in the set $F$
attains its largest value $8-2\sqrt{5}$ at the point $P=\bigl(0,0,1/\sqrt{5}\bigr)$.

$\quad$Suppose that $x\in\mathbb{Q}^3\setminus\mathbb{Z}^3$ is a zero of $f$,
and let $y:=\mathrm{round}(x)$.
If $q(x)<1$ (we certainly have $q(x)>0$) then fine,
we make a step to the next zero of $f$ with smaller denominator
this side of the Euclidean horizon.
Otherwise $q(x-y)\geq 1$, we are on the far side, and must tread more carefully.
Let $x=a/b=(a_1,a_2,a_3)/b$ be the reduced representation.
We claim that since $q(x)=m$ is an integer, the denominator $b$ is odd.
Suppose that $b$ is even;
then at least one of $a_1$, $a_2$, $a_3$ is odd,
and so in $q(x)=q(a)/b^2$ the numerator $q(a)$ is congruent to $1$, $2$, or $3$ modulo $4$,
while the denominator is divisible by $4$, contradiction.
Note that $b$ being odd implies $0 \leq \left|x_i-y_i\right| < \frac{1}{2}$, $i=1,2,3$.
Now we choose an integral point $z$, close to the integral point $y$.
If $a_1-by_1$ is even, then we set $z_1:=y_1$.
If $a_1-by_1$ is odd, we let $z_1$ be a second closest integer to $x_1$
(there are two possible choices for $z_1$ iff $x_1$ is an integer,
and we may choose either of them);
then $z_1=y_1\pm1$, $a_1-bz_1$ is even, and $\left|x_1-z_1\right|=1-\left|x_1-y_1\right|$.
In either case we have $\left|x_1-z_1\right|\leq1 - \left|x_1-y_1\right|$.
The coordinates $z_2$ and $z_3$ are chosen analogously.
We step to the next zero $x'=a'/b'$ of $f$ along the line laid through the points $x$ and $z$.
Since $\bigl(\left|x_1-y_1\right|,\left|x_2-y_2\right|,\left|x_3-y_3\right|\bigr)\in F$,
it follows that
$$q(x-z)\leq q\bigl(1-\left|x_1-y_1\right|,1-\left|x_2-y_2\right|,1-\left|x_3-y_3\right|\bigr)
\leq 8-2\sqrt{5} < 4~.$$
By the choice of the point $z$ all three coordinates of $a-bz$ are even,
thus $\mathrm{gcd}(a-bz)\geq 2$,
and Lemma 1 tells us that $b'<b$.

$\quad$Done.

For the second example we consider $q(X_1,X_2,X_3)=X_1^2+X_2^2+2X_3^2$
and $f(X)=q(X)-m$ with $m$ a positive integer.
The quadratic form $q$ is barely non-Euclidean:
for every $x\in\mathbb{Q}^3\setminus\mathbb{Z}^3$ there is $y=\mathrm{round}(x)\in\mathbb{Z}^3$
such that $q(x-y)\leq 1$.
The problem is that there exist points $x$ for which $q(x-y)=1$ is the best we can do:
if $x\in M := \mathbb{Z}^3+\bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigr)$,
then $q(x-y)\geq 1$ for any $y\in\mathbb{Z}^3$.
Note that $q(x)$ is an odd integer for every $x\in M$,
thus there do exist integers $m$, all of them odd,
so that $f$ has rational zeros,
but woe, it also has a rational zero,
with all three coordinates precisely halfway between consecutive integers,
at which we get stuck, because there is no Euclidean step from it to another zero.
On the other hand, if $m$ is even we never get stuck,
there is always a Euclidean step from a non-integral zero;
that is, though $f$ is not Euclidean, it is 'conditionally' Euclidean
on the set of its zeros.

$\quad$Now suppose that $m$ is odd and that $f$ has rational zeros,
and that we walked ourselves into a point $x=a/2\in M$.
In this case the trick we have used in the preceding example does not work,
because $\mathrm{gcd}(a-2y)^2\leq f_2(x-y)$ for any $y\in\mathbb{Z}^3$;
we must seek help from the other factor $\mathrm{gcd}(A,B,C)$
in the denominator on the right hand side of (1).
Note that we can round the point $x$ to any of the eight integral points
$x+\frac{1}{2}(\delta_1,\delta_2,\delta_3)$, where $\delta_1,\delta_2,\delta_3\in\{-1,1\}$;
let $y$ be one of these eight points.
We have $v=(\delta_1,\delta_2,\delta_3)$,
and $F(T) = f(y+Tv) = q(v)T^2 + 2\langle y,v\rangle T + q(y) - m$,
where $\langle\text{-},\text{-}\rangle$
is the bilinear form associated with the quadratic form $q$,
$\langle X,Y\rangle = X_1Y_1+X_2Y_2+2X_3Y_3$.
The leading coeffient $A=q(v)=4$ is even,
the next coefficient $B=2\langle y,v\rangle$ is also even,
thus it remains to make $C=q(y)-m$ even.
But this is easy: choose $\delta_1$ and $\delta_2$ so
that one of $y_1$, $y_2$ is even and the other one is odd.
Such a choice makes $\mathrm{gcd}(A,B,C)\geq 2$,
and since $q(x-y)=1$, we can step to a zero $x'=a'/b'$ of $f$ with $b'<2$,
that is, to an integral zero (whence $\mathrm{gcd}(A,B,C)=2$).

$\quad$Done.