Here is another way of going about the task.
The maximum value of the sum is $$1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2=204$$ and the target value is $44$ less than this.
A transposition switching $a$ and $b$ reduces the total by $$a^2+b^2-ab-ba=\frac 12\left((a-b)^2+(b-a)^2\right)[=(a-b)^2]$$
Note we want the first form here and not the simplest, as will appear later.
The $3$-cycle $(abc)$ reduces the total by $$a^2+b^2+c^2-ab-bc-ca=\frac 12\left((a-b)^2+(b-c)^2+(c-a)^2\right)$$ and $(abcd)$ by $$\frac 12\left((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\right)$$and now the pattern is clear.
So we are first looking for decompositions of $2\cdot 44=88$ into sums of up to eight squares, the greatest of which is $\leq49$, and then converting the sum of squares into a cycle decomposition, if possible.
So, for example $88=5\cdot 4^2+2\cdot 2^2$, giving us $7$ squares, which in theory allows a decomposition as a $2$-cycle and a $5$-cycle - but we can't have a $5$-cycle where all the elements have the same parity. The other alternatives are $3,4$ and $2,2,3$. The $3$-cycle is constrained to have differences $2,2,4$ and is all odd or all even - i.e. $(1,3,5), (2,4,6), (3,5,7), (4,6,8)$. The remaining part has to provide four differences equal to $4$ and this is only possible with two transpositions $(1,5)(3,7)$ if the $3$-cycle contains even numbers and $(2,6)(4,8)$ otherwise. [correction] Note that the inverses of the four three cycles also work. So this decomposition gives us eight possible permutations.
It is now a question of being systematic, as there are only a few possibilities.
Looking at a simpler example $$a+2b+3c+4d=27$$ the greatest possible sum is $1+4+9+16=30$ the difference is $3$ so we are looking at up to four squares which sum to $2\cdot 3=6$. The only possibility is $1+1+4$ representing differences of $1,1,2$.
We now need to divide this into two subsets with the same sum - $1+1=2$. This is because we start on one number, have some ups and downs and end back where we started. So the ups have to equal the downs. Three squares means a $3$-cycle. The $2$ is either $|3-1|$ or $|4-2|$. So we have the cycles $(312), (132), (423), (243)$. It depends which way you are used to reading cycles, but the solutions are then $(2,3,1,4), (3,1,2,4), (1,3,4,2), (1,4,2,3)$
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Note also that the number of cases for the original problem is intermediate for doing this by hand. The number of odd squares is a multiple of four. I get (to illustrate some of the complexities - care is required):
$7,6,1,1,1$ with $7+1=6+1+1$ representing a $5$-cycle and $7=6+1; 1=1$ representing a $3$-cycle and a transposition. The first of these requires differences of $6$ and $7$ so $8\to 1\to 7$ or $1\to 8\to 2$ or the reverse, and the differences of $1$ don't fit in to make a cycle, so you need the transposition e.g. $(34)$
$7,5,3,2,1$ with $7+2=5+3+1$ ($5$-cycle e.g. $(18354)$ - others are possible)
$7,5,3,1,1,1,1,1$ with $7+3=5+1+1+1+1+1$: $7+1+1+1=5+3+1+1$: $7+1+1=5+3+1;1=1$: $7+1=5+3;1=1;1=1$ (this is a four-cycle and two transpositions of adjacent numbers e.g. $(1856)(??)(??)$ but nothing goes with $7$ or $(1834)(??)(??)$ and nothing goes with $2$ so no solution)
$7,5,2,2,2,1,1$ with $7+2+1=5+2+2+1$: $7+2=5+2+2; 1=1$: $7+1=5+2+1; 2=2$: $7=5+2;2+1=2+1$: $7=5+2; 2=2; 1=1$
Taking $7=5+2; 2+1=2+1$ we have a 3-cycle and a 4-cycle. $7=|8-1|$ is the only possibility, so we have $(186), (183), (813), (816)$ and taking the first of these we need to realise the $1+2=1+2$ note that $7$ can't fit into this pattern , so $2$ has to go somewhere (+1 or +2) and so we get $(2354)$ or $(2453)$, and similarly for the others.
$7,4,4,2,1,1,1$ gives $7+2+1=4+4+1+1$ (from which $2=1+1$ or $1=1$ can be extracted) and $7+1+1+1=4+4+2$ e.g. for this last $(1867345)$ or $(1845673)$ (not exhaustive)
My full list is (copying screwed the spacing) - I'm adding [n] to show the number I've found of each type (corrections welcome):
$$
\begin{array}{ccccccccl}
7 & 6 & 1 & 1 & 1 & & & & \text{[16]} \\
7 & 5 & 3 & 2 & 1 & & & & \text{[10]} \\
7 & 5 & 3 & 1 & 1 & 1 & 1 & 1 & \text{[4]} \\[2ex]
7 & 5 & 2 & 2 & 2 & 1 & 1 & & \text{[28]} \\[2ex]
7 & 4 & 4 & 2 & 1 & 1 & 1 & & \text{[32]} \\
7 & 4 & 3 & 3 & 2 & 1 & & & \\
7 & 4 & 3 & 2 & 2 & 2 & 1 & 1 & \\[2ex]
7 & 3 & 3 & 3 & 3 & 1 & 1 & 1 & \\[2ex]
7 & 3 & 3 & 3 & 2 & 2 & 2 & & \\
6 & 6 & 4 & & & & & & \text{(impossible)} \\
6 & 6 & 3 & 2 & 1 & 1 & 1 & & \\
6 & 6 & 2 & 2 & 2 & 2 & & & \text{[8]} \\
6 & 5 & 5 & 1 & 1 & & & & \text{(impossible)} \\
6 & 5 & 4 & 3 & 1 & 1 & & & \\
6 & 5 & 4 & 2 & 2 & 1 & 1 & 1 & \\[2ex]
6 & 5 & 3 & 3 & 3 & & & & \text{(clearly impossible mod $3$)} \\
6 & 5 & 3 & 3 & 2 & 2 & 1 & & \\
6 & 4 & 4 & 4 & 2 & & & & \text{[16]} \\
6 & 4 & 4 & 4 & 1 & 1 & 1 & 1 & \\[2ex]
6 & 4 & 4 & 3 & 3 & 1 & 1 & & \\
6 & 4 & 4 & 2 & 2 & 2 & 2 & 2 & \\[2ex]
6 & 4 & 3 & 3 & 3 & 3 & & & \text{(impossible)} \\
6 & 4 & 3 & 3 & 3 & 2 & 2 & 1 & \\[2ex]
5 & 5 & 5 & 3 & 2 & & & & \text{[20]} \\
5 & 5 & 5 & 3 & 1 & 1 & 1 & 1 & \\[2ex]
5 & 5 & 4 & 4 & 2 & 1 & 1 & & \\
5 & 5 & 4 & 3 & 3 & 2 & & & \\[2ex]
5 & 5 & 4 & 3 & 2 & 2 & 2 & 1 & \\[2ex]
5 & 5 & 3 & 3 & 3 & 3 & 1 & 1 & \\[2ex]
5 & 4 & 4 & 4 & 3 & 2 & 1 & 1 & \\[2ex]
5 & 4 & 4 & 3 & 3 & 3 & 2 & & \\
5 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & \\[2ex]
4 & 4 & 4 & 4 & 4 & 2 & 2 & & \text{[8]} \\[2ex]
\end{array}
$$
