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
Here is one that has a complete border and insides with non-primes. (All numbers in black are composite).
$$ \begin{matrix} \ 2700 & \ 2701 & 2702 & \ 2703 & 2704 & \ 2705 & \ 2706 \\ \ 3669 & 3670 & \color{red}{3671} & 3672 & \color{red}{3673} & 3674 & 3675 \\ 4638 & \color{red}{4639} & 4640 & 4641 & 4642 & \color{red}{4643} & 4644 \\ 5607 & 5608 & 5609 & 5610 & 5611 & 5612 & 5613 \\ 6576 & \color{red}{6577} & 6578 & 6579 & 6580 & \color{red}{6581} & 6582 \\ 7545 & 7546 & \color{red}{7547} & 7548 & \color{red}{7549} & 7550 & 7551 \\ 8514 & 8515 & 8516 & 8517 & 8518 & 8519 & 8520 \\ \end{matrix} $$