At least at $p=2$, Toda's calculations do not go nearly that far up.
There are later papers that go a little further, but his ``Composition
methods in homotopy groups of spheres'' goes up to $n=19$. It is misleading
to think of $S^2$ as particularly simple. There is an old theorem (Serre?)
that if $X$ is any simply connected finite CW complex that is not contractible,
such as $S^2$, then for each prime $p$ there are infinitely many $n$ such that
$\pi_n(X)$ has $p$-torsion.
Sorry to be late to the party, but here's an example of 2 compact simply connected manifolds which have the same homotopy groups, same homology groups, same cohomology ring, and yet are not homotopy equivalent. The examples are motivated by Grigory M's examples:$S^2\times S^2$ and $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$. His examples are both $S^2$ bundles over $S^2$.
If we extend this further, it turns out there are precisely two $S^3$ bundles over $S^2$. Of course, one is the product $S^3\times S^2$, while another doesn't have a more common name, so I'll just denote it $S^3\hat{\times} S^2$.
Both of these spaces are diffeomorphic to quotients of free linear $S^1$ actions on $S^3\times S^3$. Letting $X$ denote either bundle, we have a long exact sequence of homotopy groups $$...\pi_k(S^1)\rightarrow \pi_k(S^3\times S^3)\rightarrow \pi_k(X)\rightarrow \pi_{k-1}(S^1)\rightarrow ...$$
which can be used to show that $\pi_k(X) \cong \pi_k(S^3\times S^3)$ for $k\geq 2$ and $\pi_1(X) = \{e\}$ and $\pi_2(X) \cong \mathbb{Z}$.
The Hurewicz theorem together with Universal coefficients theorem implies $H^1(X) = 0$ and $H^2(X) \cong \mathbb{Z}$. Poincare duality then forces the rest of the cohomology rings to agree.
Finally, to see $S^3\times S^2$ and $S^3\hat{\times} S^2$ are different, one computes the Stiefel-Whitney classes of their tangent bundles. It turns out $w_2(S^3\times S^2) = 0$ while $w_2(S^3\hat{\times}S^2)\neq 0$. (And all other Stiefel-Whitney classes are $0$ for both spaces). Since the Stiefel-Whitney classes can be defined in terms of Steenrod powers, they are homotopy invariants, so $S^2\times S^3$ and $S^3\hat{\times}S^2$ are not homotopy equivalent.
Best Answer
I have come up with a counterexample:
$S^2$ and $S^3 \times \mathbb{C}P ^\infty$ have the same homotopy groups, but their suspensions don’t since $\pi_7 (S^3)$ is torsion but $\pi_7(S(S^3 \times \mathbb{C}P ^\infty))$ contains an infinite element since $\pi_7(S^4)$ is not torsion.