Since the Guralnick and Tiep paper is very long, I thought I would summarize my understanding of it. Note that I only learned about this result from moonface last night, so I am hardly an expert. This answer is community wiki, in case anyone can improve on my summary.

We want to establish the following result: Let $G$ be a noncommutative subgroup of $GL(V)$, with $V$ a $\mathbb{C}$ vector space. Let $k \geq 6$. Then $\mathrm{Sym}^k(V)$ is reducible. (If $G$ is commutative and $V$ is one dimensional then, of course, $\mathrm{Sym}^k(V)$ is always one dimensional and hence irreducible.)

Our proof is by induction on $|G|$. Our base case will be when $G$ is a central extension of a simple group.

Choose a nontrivial normal subgroup $H$ of $G$ such that $G/H$ is simple. If we can't do this then $G$ is simple and we are in the base case. Let $V \cong \bigoplus U_i$ be the decomposition of $V$ into $H$-isotypic components. If there is more than one summand, then $\bigoplus \mathrm{Sym}^k(U_i)$ is a nontrivial $G$-subrep of $\mathrm{Sym}^k(V)$. So we may assume that $V \cong W \otimes X$, where $W$ is an $H$-irrep and $H$ acts trivially on $X$. Then one can show (proof of lemma 2.5) then $G$ is contained in $GL(W) \times GL(X)$. (This is nontrivial but not deep; you could give it as a problem in a graduate-level representation theory course.)

Then $\mathrm{Sym}^k(W) \otimes \mathrm{Sym}^k(X)$ is a subrepresentation of $\mathrm{Sym}^k(V)$. This subrep is proper unless $\dim X=1$ or $\dim W=1$. If $\dim X=1$, then $V$ is an irrep of $H$ and $\mathrm{Sym}^k(V)$ is irreducible as an $H$-rep, so we are done by induction.

If $W$ is one dimensional, then $H$ acts on $V$ by scalars, so $H$ is central. So we are in our base case: a central extension of a simple group. This case is done by group cohomology and the classification of simple groups. See section 4 for groups of Lie type, section 6 for alternating groups and section 7 for sporadic groups.