So, if you make compact analytic varieties algebraically, you can'tif you make compact analytic varieties algebraically, you can't escape from the algebraic class some of the main constructions of complex manifolds do not escape from the algebraic class. All (But not all: deformations and infinite group actions can escape.) All projective analytic varieties are algebraic, and in dimension 1 all compact curves are projective. Moreover, there are limited ways for a compact analytic manifold to avoid being projective, by Moishezon's theorem and Kodaira's theorem. In practice, then, most of the complex manifolds that people make are algebraic. Also, most of the analytic calculations on a proper algebraic variety are algebraic: Many global calculations are algebraic by GAGA, and many local calculations are algebraic just by truncating Taylor series.
Contrast all this with real algebraic vs real analytic. It is still true that (the real points of) a smooth real algebraic variety is a real analytic manifold. More strongly than in the complex case, although it is highly non-trivial, every compact real analytic manifold is real algebraic. But the real algebraic structure is massively not unique, even for a circle, and that makes all the difference.
The other answerers in this thread, who are more expert in this topic than I am, had more information about why a compact complex manifold might not be a smooth projective variety. Just for clarity, I will restrict attention to the compact, smooth case. Also, you say "proper" rather than "compact" in the algebraic category because every algebraic variety is "compact" in the extremely coarse Zariski topology. An algebraic variety is proper if and only if it is analytically compact. The main use of the word proper is to emphasize that it is more general than projective, which means given by polynomial equations in complex projective space.
There are two very different initial reasons that an analytic complex manifold might not be projective. It might not be Moishezon: A complex $n$-manifold is Moishezon if it has $n$ algebraically independent meromorphic functions. (The number of algebraically independent elements or the transcendence degree of a field is called the Krull dimension. The meromorphic Krull dimension of a compact complex $n$-manifold is at most $n$.) Or it might not be Kähler: A complex $n$-manifold is Kähler if it has a Riemannian metric such that the covariant derivative of the complex structure vanishes. So to summarize what people said about compact complex manifolds (much of which is in the back of Hartshorne's book):
projective ⇒ algebraic ⇒ Moishezon ⟺ bimeromorphically projective
projective ⇒ Kähler ⇒ symplectic ⇒ non-zero $H^2$
algebraic ⇒ non-zero $H^2$ (exposited by David Speyer)
Moishezon and Kähler ⟺ projective (Moishezon)
Kähler and integrally symplectic ⟺ projective (Kodaira)
In addition, projective and algebraic structure and the Moishezon property are all unstable with respect to analytic deformation. And bimeromorphic equivalence preserves $\pi_1$. Taubes found compact complex manifolds that have the wrong $\pi_1$ to be Kähler; indeed they can have any $\pi_1$. Voisin found compact Kähler manifolds with the wrong homotopy type to be projective, disproving Kodaira's conjecture that every compact Kähler manifold can be deformed to projective.
Still, despite these beautiful ways to make complex manifolds that aren't projective, it's generally easier to study projective examples. It's generally easier to sidestep analysis and do algebra instead.