Use the initial conditions like this:

$\displaystyle \begin{align*} y(x) &= \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \ldots \\ y(0) = 1 &= a_0 \end{align*}$

and similarly for $\displaystyle y'(0)= -1$ to fix $\displaystyle a_1$.

Then put the series for $\displaystyle y(x)$, $\displaystyle y'(x)$ and $\displaystyle y''(x)$ into the differential equation and compare coefficients of $\displaystyle x^n$. You can start with coeffients of $\displaystyle x^2$ if you like for which you will get an expression relating $\displaystyle a_2$ to $\displaystyle a_0$ and $\displaystyle a_1$ which you already know. You can then continue with coefficients of $\displaystyle x^3$, $\displaystyle x^4$, etc.. But if you need a general expression for the $\displaystyle a_n$ in terms of $\displaystyle n$, you will have to compare coefficients of $\displaystyle x^n$. This gives a relation between $\displaystyle a_n$, $\displaystyle a_{n-1}$ amd $\displaystyle a_{n-2}$. This might be something like the recurrence relation for the Fibonacci numbers $\displaystyle a_n=a_{n-1}+a_{n-2}$ which can be solved (using the values of $\displaystyle a_0$ and $\displaystyle a_1$) to get a closed form expression $\displaystyle a_n=f(n)$.