profinite completion and linear representations of finitely presented groups
Yes, there exists such a finitely presented group.
Let $\Gamma$ be a cocompact arithmetic lattice in a product of $\ge 2$ rank 1 groups simple groups (with trivial center) over locally compact fields of finite characteristic. So $\Gamma$ is finitely presented (it is even CAT($0$)). By Malcev, $\Gamma$ is residually finite. By Kazhdan-Margulis, $\Gamma$ is just-infinite: all normal subgroups in $\Gamma$ except $\{1\}$ have finite index. By Venkataramana's superrigidity, every finite-dimensional complex representation of $\Gamma$ has a finite image.
My initial answer was conditional answer:
Let $\Gamma$ be a cocompact lattice in $\mathrm{PSp}(n,1)$, $n\ge 2$. This is a non-elementary hyperbolic group. Hence it is known that for many $x\in\Gamma-\{1\}$, the quotient $\Lambda_x=\Gamma/\langle\!\langle x\rangle\!\rangle$ is non-elementary hyperbolic.
It is also known (superrigidity) that every linear proper quotient of $\Gamma$ is finite. Hence, every linear quotient of $\Lambda_x$ has a finite image. Accordingly
- either $\Lambda_x$ has an infinite profinite completion, hence answers your question
- or $\Lambda_x$ has finite profinite completion, and hence there exists a non-residually-finite hyperbolic group. The existence of such a group is a famous open problem.
I'm not sure that any well-accepted conjecture predicts which of these properties holds (this might even depend on $x$).
Yes if you weaken to finitely generated. Take an infinite finitely generated torsion group that is residually finite like the Grigorchuk group. By a theorem is Schur any finitely generated torsion linear group is finite. I am not sure of the general case.