Can finitely many values of a polynomial determine it? Let $d$ be a positive integer greater than 2. Define an equivalence relation on monic integer polynomials of degree $d$: $f\sim g \iff f(k_1 x+k_2)=g(k_3 x+k_4)$ for some integers $k_1,...,k_4$.

Is there a number $m$ such that for any $m$ distinct integers, there is at most one equivalence class that attains these at integer coordinates?

I ask for $d>2$ since it is vacuous for $d=1$ (only one class) and it fails for $d=2$: $x^2-1$ and $2y^2$ are not in the same class and having infinitely many common values.
From some short calculations I think it is true that for $d+1$ (maybe a bit more) distinct values there are at most a finite number of equivalence classes possible, but I don't see how to bound the number of classes uniformly, let alone by 1.
 A: You are asking for a condition on $f$ and $g$ such that the curve:
$$C: f(y) - g(x) = 0$$
has a uniformly bounded number of rational points.
Note that if $f$ and $g$ are equivalent under an affine transformation,
then $C$ is divisible by a linear factor and is not reducible. The converse is
almost true. Namely, as long as the degree of $f$ and $g$ is sufficiently
large, and $f$ and $g$ are not of the form
$a \circ b$ for polynomials $a$ and $b$ of degree $> 1$,
then $C$ will be irreducible. (This follows from CFSG. Of course, using composition of functions
one can create many degenerate examples: $P(y)^2 - Q(x)^2$ is divisible
by $P(y) - Q(x)$. The example in the comments giving a example
in even degrees arises in this way, by taking a degree two example
and using composition.)
If the degree $d$ of $f$ (and $g$) is prime, then $f$ and $g$
are certainly indecomposable, so let's concentrate on that case, since there
are no reductions to smaller degree. For convenience, let's also only consider
the case when $C$ is irreducible (if $d$ is prime, this is automatic if $d$ is
sufficiently large, by the remark above.)
If $C$ has genus at least two, then $C$ will have only finitely
many rational points (Faltings). Work of
Caporaso, Harris, and Mazur:
http://www.ams.org/journals/jams/1997-10-01/S0894-0347-97-00195-1/home.html
suggests that the number of solutions may be even be bounded in terms of the genus, and hence in terms of the degree.
Whether you believe Lang's conjectures or not, you are unlikely to disprove
Lang's conjectures easily, so any negative example to your claim should come from 
a pair of functions $f$ and $g$ so that $C$ has small genus.
In small genus, we may have many rational points, but as far as integral points
we also have Siegel's theorem to content with. A projective model
$\widetilde{C}$ of $C$
is given by $Z^d f(Y/Z) - Z^d g(X/Z) = 0$. Setting $Z = 0$, we obtain the equation
"at infinity"
$Y^d - X^d = 0$, which has $d$ points over the complex numbers.  Hence, assuming $d \ge 3$,
$$\# \widetilde{C} \setminus C \ge 3.$$
 By Siegel's theorem we deduce 
$C$ has only finitely many integral solutions.  Thus, when the degree $d$ is
prime and sufficiently large (or more generally, providing one avoids degeneracies arising from the phenomena alluded to in the first paragraph), any $f$ and $g$ in different equivalence classes
will only coincide on a fixed number of integers.
Your question, however, asks whether there is a uniform bound. There
is certainly no uniform bound for Siegel's theorem, at least when the genus
is $\le 1$.
 There is a standard “renomalization” trick which takes
a curve with infinitely many rational points and produces a curve with many
integral points. This trick works in this case. 
Specifically, suppose that $C: f(y) - g(x) = 0$ has infinitely
many rational points. Then there certainly exists some integer 
 $N$ such that $C$
has a bizillion points of the form $(u/N,v/N)$ (take $N$ to be a common denominator).  We may then write down the different integral
model:
$$C': N^d f(y/N) - N^d g(x/N) = 0,$$
which now has a bizillion integral points $(u,v)$. This also allows one to answer your question in general degrees, simply by choosing $f$ and $g$ so that $C$ has
infinitely many rational points, and then renomalizing appropriately.
The easiest specific example would be to take $C$ of genus zero. 
For example, take $f = t^n$ and $g = t^{n-1}(t-1)$.
Then 
$C: f(y) - g(x) = y^n - x^{n-1}(x-1)$ has genus zero, as can be seen from the 
parametrization
$$x  = \frac{1}{1 - t^n}, \qquad y = \frac{t}{1 - t^n}.$$
From the above construction, there will exist positive integers $N$ such that
the polynomials $t^n$ and $t^{n-1}(t - N)$ will take on the same bizillion values.
This answers your question in the negative.
EDIT: I guess the last example can be made quite concrete.
Let 
$$N = (1 - 2^d)(1 - 3^d)(1 - 4^d) \ldots (1 - M^d).$$
Then $t^d$ and $t^{d-1}(t-N)$ both take on the values
$\displaystyle{\left(\frac{aN}{1 - a^d} \right)^d}$
for $a = 2, \ldots, M$.
A: This is not true. $f(x)$ and $f(2x)$ will not be in the same equivalence class in general, yet their images will agree on infinitely many points. Hence there is no fixed value $m$ such that any $m$ points determine a single equivalence class 
