A hyperplane is a hypersurface cut out by a single linear polynomial, thus a hypersurface of degree 1.
I have found the following mentionings of curves (in Hartshorne):
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An affine curve is defined to be $f(x,y)=0$ for $f$ an irreducible polynomial.
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A curve is a variety of dimension 1.
Using (1), a hyperplane is a special circumstance of a curve, since $f$ being linear implies that $f$ is irreducible.
Using (2) this implies that all hyperplanes are varieties of dimension $1$.
Moreover, this would imply that hyperplanes have degree and dimension equal to $1$.
However, in exercise 2.8 Hartshorne asks us to prove that a projective variety $Y \subset \mathbb{P}^n$ has dimension $n-1$ iff it is the zero set of a single irreducible homomogeneous polynomial of positive degree. We call such varieties hypersurfaces.
A hyperplane is a hypersurface and thus must have dimension $n-1$ by the above statement.
A hyperplane can also be considered a curve and thus must have dimension $1$.
These statements only work together when $n=2$.
Is it wrong to consider a hyperplane as a special type of curve in $\mathbb{P}^3$?
Something is going wrong in my implications since it only works out for $n=2$.
Best Answer
As Nesos' comment says, curves are any 1 dimensional objects, while hypersurfaces are codimension 1 objects - given by the vanishing of one equation. When ambient space is 2 dimensional (i.e. $\mathbb{A}^2$ or $\mathbb{P}^2$), hypersurfaces are curves, and indeed curves are hypersurfaces. But in higher dimensions, hypersurfaces are not curves.
Lubin's comment reminded me, hyperplanes are hypersurfaces given by the vanishing of a single linear polynomial (regardless of the dimension of the ambient space). So in $\mathbb{P}^3$, hyperplanes would be isomorphic to $\mathbb{P}^2$, and would not be curves.