I want to show that the diophantine equation does only have the trivial solution $x=y=0$. Since $\text{gcd}(73,-137)=1|0$ this is solveable. So
\begin{align} 137&= 1\cdot73+64\\ 73 &= 1\cdot64+9\\ 64 &= 9\cdot7+1\\ 7 &= 1\cdot7+0 \end{align}
So,
\begin{align} 0&=7-7\cdot1=7-7(64-9\cdot7)=7-7\cdot64+49\cdot9=7-7(137-73)+49(73-64)\\ &=7-56\cdot137+105\cdot73. \end{align}
But this is not in the correct form, the $7$ at the start bothers me and I cant get rid of it without ruining my $137$ or $73$.
How can I fix this?
The equation can be written $73x=137y$. This means that this number is a common multiple of $x$ and $y$, so it is a multiple of their lowest common multiple, which is $73\cdot137$, because $\gcd(73,137)=1$.
Thus there exists $n$ with $73\cdot137n=73x=137y$. Thus $x=137n$, $y=73n$. Conversely, any pair of numbers of this form is a solution.
When you apply the reverse Euclidean algorithm, you don't start from the line with the zero remainder, but from the line with the least nonzero remainder: \begin{align} 1&=64-9\cdot7\\ &=64-(73-64)\cdot7=73(-7)+64\cdot8\\ &=73(-7)+(137-73)\cdot8\\ &=137\cdot8-73\cdot15 \end{align}