There are some more-or-less equivalent conditions:

Bass--Serre theory says that a group is freely indecomposable if and only if it acts on a tree with trivial edge stabilizers and no global fixed point.

More deeply, Stallings' Ends Theorem asserts that a finitely generated group splits over a *finite* subgroup if and only if it has more than one end.

As these are equivalent (modulo finite=trivial), you might say that they're not very interesting. Truly interesting necessary conditions are rather hard to write down, since being freely indecomposable is, in a sense, generic.

Here's one, much stronger, sufficient condition:

If a group $\Gamma$ has a finite generating set $S$ such that every element of $S$ is torsion and every element of $S^2$ is torsion then $\Gamma$ doesn't split at all, let alone freely.

This is essentially a consequence of Helly's Theorem for trees.

Here's an outline of a proof. Suppose that $\Gamma$ acts on a tree $T$. We want to prove that $\Gamma$ fixes a point.

- Prove that if $\gamma\in\Gamma$ has finite order then $\gamma$ fixes a point.
- Prove that if the sets of fixed points of $\gamma,\delta\in\Gamma$ are disjoint then $\gamma\delta$ has infinite order. You can do this by constructing an isometric copy of $\mathbb{R}$ in $T$ that $\gamma\delta$ translates.
- Prove Helly's Theorem for trees: if $\{T_i}$ is a finite set of pairwise-intersecting subtrees of $T$ then $\bigcap_i T_i\neq \varnothing$.
- Conclude as follows. Because every $s\in S$ is torsion, each such $s$ has a fixed point. Because $st$ is torsion for $s,t\in S$, the sets of fixed points pairwise intersect. By Helly's Theorem, there is a point fixed by every $s\in S$, and hence by $\Gamma$.