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.
You're correct this is related to the fundamental theorem of arithmetic. The statement regarding a powerful or squarefull (as Gerry Myerson's question comment suggests naming them instead of fortified) number means all of its prime factors must have a power of $2$ or larger.
You're asking to prove that all such numbers $n$ can be expressed using natural numbers $a$ and $b$ where
$$n = a^2b^3 \tag{1}\label{eq1A}$$
This involves basically proving that for every integer $m \ge 2$, there are non-negative integers $c$ and $d$ such that
$$m = 2c + 3d \tag{2}\label{eq2A}$$
You state in your comment you can do this, so I'll leave that to you, as it's fairly easy to do, such as with induction. Also note it's related to a particular case of the coin problem which shows the maximum number that can't be expressed by \eqref{eq2A} is $a_1 a_2 - a_1 - a_2 = 2(3) - 2 - 3 = 1$, so anything $2$ or larger can.
With this, consider each prime factor $p$ of $n$. Consider the p-adic valuation function $v_p(m) = e$ which gives, for any prime $p$, the highest exponent $e$ such that $p^{e} \mid m$. Then for any prime factor $p$ of $n$, have $v_p(n) = m$, with $v_p(a) = c$ and $v_p(b) = d$. Do this for all $p \mid n$ to create the corresponding prime factors of $a$ and $b$, and with both $a$ and $b$ not having any other prime factors. You will then end up with \eqref{eq1A}.
Best Answer
If you let $a_n = 1$ if $n$ has this property and $a_n = 0$ otherwise, then
$$F(s) = \sum \frac{a_n}{n^s} = \prod_{p} \left(1 + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \frac{1}{p^{5s}} + \ldots \right),$$
On the other hand,
$$\zeta(2s) = \prod_{p} \left(1 - \frac{1}{p^{2s}}\right)^{-1},$$
and thus
$$\frac{F(s)}{\zeta(2s)} = \prod_{p} \left(1 + \frac{1}{p^{3s}} - \frac{1}{p^{4s}} - \frac{1}{p^{9s}} + \ldots \right),$$
where the exponents are those occuring in $(1-x^2) \sum x^p = 1 + x^3 - x^4 - x^9 + \ldots$.
From this, you see that $F(s)$ is holomorphic up to $s = 1/2$ where there is a simple pole with residue $C/2$, where
$$C = \prod_{p} \left(1 + \frac{1}{p^{3/2}} - \frac{1}{p^{2}} - \frac{1}{p^{9/2}} + \ldots \right),$$
$$ = 1.4310606003 \ldots $$
But then, the Wiener–Ikehara theorem (and its natural variants)
$$\sum_{n<X} a_n \sim C x^{1/2} + O(x^{1/3}).$$