If for $A^2$ the problem is, as you have said, "pretty easy" then for any $A^k$...
The proof can be made without referring to eigenvalues - it is possible to prove the claim only with the application of basic definitions for symmetric matrix and nilpotent matrix)
Note that for $A$ which is symmetric matrix $(A^2)^T= (AA)^T=A^T A^T =(A^T)^2=A^2$ so square of symmetric matrix is also symmetric.
You have proved, I assume*, that for any non-zero symmetric matrix $A^2 \ne 0 $.
*(indeed it's easy to prove than there are some non-zero values at least on the main diagonal.
Denote symmetric matrix as matrix consisting of columns
$A=\begin{bmatrix}
v_{1} & v_{2} & \dots & v_{n} \\ \end{bmatrix}$
Then $A^2=AA=A^TA=\begin{bmatrix}
v_{1}^T \\ v_{2}^T \\ \dots \\ v_{n}^T \\ \end{bmatrix}\begin{bmatrix}
v_{1} & v_{2} & \dots & v_{n} \\ \end{bmatrix}= \begin{bmatrix}
v_{1}^Tv_{1} & v_{1}^Tv_{2} & \dots & v_{1}^Tv_{n} \\ v_{2}^Tv_{1} & v_{2}^Tv_{2} & \dots & v_{2}^Tv_{n} \\\dots & \dots & \dots & \dots \\ v_{n}^Tv_{1} & v_{n}^Tv_{2} & \dots & v_{n}^Tv_{n} \end{bmatrix} $
but $ v_{i}^Tv_{i} =\Vert v_i \Vert ^2 $ hence on the main diagonal are squares of column vectors lengths. At least one of these vectors $v_i$ is non-zero vector).
If $A^2 \ne 0 $ then the same can be said for $(A^2)^2$, $((A^2)^2)^2$, etc..
Then always it is possible to find for any $k$ the number $ m > k $ that $A^m \ne 0$ with the application of iterative squaring the result matrix as above..
But $A^m=A^{m-k}A^k$.
Because $A^m \ne 0$ so $A^k$ can't be equal $0$, otherwise $A^m$ should have been equal $0$ as well.
So we have that for any $k<m$ ($m$ can be made arbitrarily big) - it is impossible for non-zero symmetric matrix $A$ to obtain $A^k=0$.
