With some more effort, I got one solution:

642793581
713856429
895241637
287165394
469328715
531479268
926517843
374982156
158634972

With some more effort, I got one solution:

642793581
713856429
895241637
287165394
469328715
531479268
926517843
374982156
158634972

I'm almost certain that this is the unique solution - not totally certain because my sketch is a bit chaotic and there could be error(s) in the deduction.

There is not so much to say about my method. As some comments suggest, it is very hard to solve without a computer, and I also used the help of a computer.

More specifically, I did everything in Photoshop. With the ability of creating layers and changing colors, I can easily go back to the last "guess point", as soon as a contradiction occurs.

The rest is just trial-and-error, plus some patience.

Partial answer (or rather, first step):

In the third row, the sum of the three middle squares is $45 - 22 - 16 = 7$.
This can only be $1 + 2 + 4$.

I claim that the right one (i.e. the sixth column) must be $1$.

It cannot be $4$, otherwise the sum of the two squares above it will be $10 - 4 = 6$, and $6$ cannot be a sum of two different numbers other than $1, 2$.
Neither can it be $2$, otherwise the two squares above it must be $3, 5$. But then the last three squares in the sixth column sum up to $13$ and none of them is $2, 3, 5$. This is impossible.

Therefore the square in the third row and sixth column is $1$. As a result, the two squares above it are $3, 7$; the two squares to the left of it are $2, 4$; the last three squares in the sixth column are $2, 5, 6$; the three middle squares in the sixth column are $4, 8, 9$.

At this point my spare time is up and I cannot proceed further...

With some more effort, I got one solution:

642793581
713856429
895241637
287165394
469328715
531479268
926517843
374982156
158634972