Firstly, we should do some preparatory work to make the system look nicer.
$$\begin{align}
\tag1(n-1)A_{n+1}-(n+2)A_n&=-(2n+1)A_1\\
\tag2(n+1)A_{n+1}-nA_n-2[A_{n+1}+A_n]&=-(2n+1)A_1
\end {align}$$
Already, at this stage, the RHS, with the constant $A_1$, is a polynomial. This suggests to us to try a polynomial for the entire expression. Note that a polynomial terminates. If we try low ordered polynomials and yet fail, we might try a formal power series.
However, we do a trick first, to expose the (anti-)symmetry. Sub $n\to-n-1:$