I can prove a very narrow form.
Let's consider numbers of the form:
$(D_{n-1}D_{n-2}.....D_{2}D_{1}D_0)_{10} = (D_{n-1}D_{n-2}.....D_2[A..F])_{16}$.
Here 2 least significant digits in decimal representation change
to [A..F]. For these numbers, conditions that need to satisfy are,
$100*x + 10 = n$ ....(1)
$16*y + 10 = n$ ....(2)
So,
$y = 6.25*x$ ....(3)
Let's assume,
$x_{10}$ is of form $....N_{k-1}N_{k-2}.....N_{2}N_{1}N_{0}$
or,
$x_{10} = ....+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ....+ (N_{2}10^2) + (N_{1}10^1) + N_{0}$
So,
$y_{10} = ....+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ....+ (N_{2}16^2) + (N_{1}16^1) + N_{0}$
Also from (3),
$y_{10} = 6.25(....+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ....+ (N_{2}10^2) + (N_{1}10^1) + N_{0})$
So,
$(....+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ....+ (N_{2}16^2)
+ (N_{1}16) + N_{0}) =
6.25(....+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ....+ (N_{2}10^2)
+ (N_{1}10^1) + N_{0})$
Using up to 6 digits for x,
$(1048576N_5 + 65536N_4 + 4096N_3 + 256N_2 + 16N_1 + N_0) =
(625000N_5 + 62500N_4 + 6250N_3 + 625N_2 + 62.5N_1 + 6.25N_0)$
or,
$42357600N_5 + 303600N_4 - 215400N_3 - 36900N_2 - 4650N_1 - 525N_0 = 0$
Value in Hex falls behind till N3 because initial modulus was 100 for decimal
and 16 for hex. But at and after N4, hex value overtakes decimal forever.
Only solution for N5 (and above) between 0 and 9 is 0. Also, there are no solutions possible less
than 5 digits for x in (1).
So essentially numbers of these forms are only 7-digits or 2 digits,
And only possible solutions are,
10-15
4494410-4494415
5660810-5660815
6784010-6784015
7950410-7950415
This is not an answer to the question (which I think is well-addressed by Micah's comment above) but a compendium of miscellaneous observations.
First, Micah's comment points out that it would be very surprising if there were not a solution for all sufficiently large $N$, and computer search bears this out: there are solutions for $N=1,15,16,17, 23,$ and all numbers between $25 $ and $50$, at which point I stopped checking. As $N$ increases, the number of solutions increases very rapidly; there is 1 solution for $N=15$, ten solutions for $N=25$, and $17,175$ for $N=35$. The $N=15$ case is atypically constrained.
Finding the solution when $N=15$ is very easy. One can begin by observing that $9$ is adjacent only to $7$, and $8$ only to $1$, so the solution, if it exists, must begin with $9$ and end with $8$. (Or vice-versa, giving the same solutions in reverse, which we henceforth disregard.) But the moves from $9$ are forced: $9-7-2-14-11-5-4-12-13-3$ is the only sequence possible. From $3$ one can go to $1$ or to $6$, and since going to $1$ evidently doesn't work (since we know $1$ is next-to-last) it must end $3-6-10-15-1-8$ and that is the only solution.
Consider the graph whose nodes are $\{1,\ldots, 15\}$ in which has two nodes are connected whenever their sum is a square. A solution to the problem is exactly a hamiltonian path in this graph. When we look at this graph, the uniqueness of the solution is completely obvious:
and even a child can see that there is only a single hamiltonian path. (I know because I checked with my six-year-old daughter, who agrees.) The graphs for $N=16$ and $N=17$ are similarly trivial, and a glance at the graph for $N=18$ or $N=19$ shows why there is no solution for those values:
In $N=20, 21, 22$ the lack of solutions is still easy to see, although the troublesome dead end at 16 has been connected to 5 via 20. For $N=24$ I did not see any obvious reason why there is no solution, but I think a simple argument could probably be made involving the disposition of $11, 22, $ and $23$.
I have not done much investigation of the analogous problem where two numbers are connected if their sum is a perfect cube. For $N=100$ the graph is not even connected. For $N=200$ it is connected, but has many leaves. Even for $N=300$ I suspect there is a fairly easy proof that there is no solution, involving the relatively independent cluster of $\{29, 35, 90, 126, 217, 253, 259\}$ which has only 4 connections to the rest of the graph.
Best Answer
The answer is: NO, the list is not complete, not even close.
$9.654.499.999^3=899889864798996889278963499999\implies 226/30=7.53$
$14.422.489.115^3=2999995888877979877897997595875\implies 233/31=7.52$
$17.087.193.298^3 = 4988984988499998899898893979592\implies 235/31=7.58$
$17.851.232.683^3=5688589987988995688999787955987\implies 235/31=7.58$
$18.751.984.083^3=6593889683979798999578997899787\implies 234/31=7.55$
$19.998.639.899^3=7998367989789967795589578889699\implies 233/31=7.52$
$20.405.777.229^3=8496878797688887969682889979989\implies 234/31=7.55$
$21.253.161.559^3=9599986686699689764859698999879\implies 235/31=7.58$
...
I think I will stop here...