Definition by Mumford's red book:
An affine variety is defined as an irreducible algebraic subset of $\mathbb{A}^n$ with a sheaf of rings of $k$-valued functions.
A projective variety is defined as an irreducible algebraic subset of $\mathbb{P}^n$ with a sheaf of rings of $k$-valued functions (page 28).
A variety is defined as a separated prevariety (page 37) and a prevariety is a locally ringed space which is locally isomorphic to an affine variety as locally ringed space.
Question: If we have a non-projective irreducible variety, then is it possible that it is topologically homeomorphic to a closed subset of $\mathbb{P}^n_k$ in the Zariski topology?
Alex's answer in the comments probably suffices, but I'll write one here. As mentioned above, $\Bbb{A}^1_k\cong \Bbb{P}^1_k$ as topological spaces with their Zariski topologies. Actually, the situation is even worse. Any pair of (quasi-compact) curves is homeomorphic because their point sets have the same cardinality and the Zariski topology coincides with the finite complement topology. In particular, given any curve in $\Bbb{P}^n_k$, it is homeomorphic to $\Bbb{A}^1_k$ and we see that your question has an affirmative answer.
As a sidenote: even if we consider curves over $\Bbb{C}$ in the classical topology, smooth curves of genus $1$ (elliptic curves) are all diffeomorphic to a torus, but this does not mean that we regard them as the same in algebraic geometry.