Both quantizing a "particle" and quantizing a "field" follow the same, completely general ideas:
From the Hamiltonian point of view, we start with a classical phase space (finite-dimensional for a "particle", infinite-dimensional for a "field"), and quantization means essentially implementing the classical Poisson algebra on the phase space as the operator algebra of a Hilbert space such that the commutator of the quantum operators coincides with the classical Poisson bracket. That is, to every classical phase space function $f$ we associate an operator $\hat{f}$, and we would like that $[\hat{f},\hat{g}] = \mathrm{i}\hbar\{f,g\}$ and $p(\hat{f}) = \widehat{p(f)}$ for any polynomial $p$.1
For the finite-dimensional case, the Stone-von Neumann theorem guarantees that the canonical Poisson bracket $\{x,p\} = 1$ leads to the unique representation of $\hat{x}$ as multiplication and $\hat{p}$ as differentiation in the position basis, but this fails for the infinite-dimensional case, which makes the canonical quantization of fields more difficult, and their analogue of wavefunctions - so-called wavefunctionals - less useful than in the particle case.
Nevertheless, quantizing particles and quantizing field is not so different - every phase space coordinate $x,p$ is promoted to a quantum operator $\hat{x},\hat{p}$, just like every field $\phi(x)$ and its canonically conjugate momentum density $\pi(x)$ are promoted to operator-valued functions $\hat{\phi}(x),\hat{\pi}(x)$ (or, more properly, operator-valued distributions) whose commutation relation is their classical Poisson bracket.
It's "just" that the infinite-dimensional phase space gives much more trouble than the finite-dimensional case, so quantizing fields is "harder".
Another way to see that quantizing fields and particles is not really different is to observe that if one thinks of the position of a particle as $x(t)$, this is a field on a 0+1 dimensional spacetime, and then usual QM becomes just the special case of a 0+1 dimensional QFT, which is best seen in the path integral approach.
1In general, this is not possible due to the Groenewold-van Hove theorem. This leads to using the Moyal bracket instead of the Poisson bracket in the deforomation quantization approach.