Aaron's answer points out the problem you had. There's a general method, too. This is a simple example of a cryptarithm, where some or all the digits in an arithmetic expression are replaced by unknown variables. For addition problems, you can write each digit sum as one or more equations (since you won't necessarily know the carry values) and solve the set of equations.
For your problem, a clever solver might first check whether there's a possible pattern: could it be that the terms being added had a particularly nice forms, like perhaps
Well, well, well. In octal, $4567+2345=7134$ and with $x=4, y=3$ we see that we've found a solution.
Of course, it won't always be that easy, but it's still not terrible to find the solution in this case. Given
we could look at the 1s-place sum: $7 + 5 = x$. Aha! in octal $7+5=4$ with a 1 carry, so $x=4$. We're halfway there, with a sum that's now
Now look at the 8's column in the sum. Recalling that the 1's column gave us a carry of 1 we have $1+6+4=y$ so in octal $y=3$. We're done, modulo checking that the 64s column sum is correct, which it is.
In a true cryptarithm, often all of the numbers are unspecified, like this:
(which does have solution(s) in octal).