Yes, they are different inner products. In fact, because there are two different conventions for how the wedge product relates to the tensor product, there are actually three different possible inner products on differential forms.
1. The Pointwise Hodge Inner Product: This is the inner product on differential forms that satisfies
$$
\langle\varepsilon^{i_1}\wedge\dots\wedge\varepsilon^{i_k},\varepsilon^{i_1}\wedge\dots\wedge\varepsilon^{i_k}\rangle = 1
$$
whenever $i_1<\dots<i_k$ and $(\varepsilon^1,\dots,\varepsilon^n)$ is a local orthonormal frame for $T^*M$. It's also characterized by
$$
\omega \wedge *\eta = \langle\omega,\eta\rangle dV_g,
$$
where $*$ is the Hodge star operator and $dV_g$ is the Riemannian volume form. In particular, on $\mathbb R^2$ with the Euclidean metric, it satisfies
$$
\langle dx^1\wedge dx^2,dx^1\wedge dx^2\rangle = 1.
$$
This inner product is universally used in Hodge theory.
(The terminology "pointwise Hodge inner product" is not standard, but it's the most accurate name I can think of for this inner product, given that the term Hodge inner product usually refers to the corresponding global inner product on differential forms obtained by integrating this pointwise inner product.)
2. The Tensor Inner Product: This is obtained from the usual inner product on tensors, by viewing the bundle of differential $k$-forms as a subbundle of the bundle of covariant $k$-tensors. The relationship between this and the pointwise Hodge inner product depends on which convention is used for wedge products. In my books, I refer to these conventions as the determinant convention and the Alt convention.
2a. Using the determinant convention:
This is the wedge product convention that satisfies
$$
\omega^1\wedge\dots \wedge \omega^k(v_1,\dots,v_k) = \det (\omega^j(v_k))
$$
when $\omega^1,\dots,\omega^k$ are covectors and $v_1,\dots,v_k$ are vectors.
In particular, with this convention, we have $dx^1\wedge dx^2 = dx^1\otimes dx^2 - dx^2 \otimes dx^1$, so
$$
\langle dx^1\wedge dx^2,dx^1\wedge dx^2\rangle = 2.
$$
2b. Using the Alt convention:
With this convention, the wedge product is determined by
$$
\omega\wedge\eta = \operatorname{Alt} (\omega\otimes \eta)
$$
when $\omega$ and $\eta$ are any differential forms whatever.
In particular, with this convention, we have $dx^1\wedge dx^2 = \tfrac 1 2 (dx^1\otimes dx^2 - dx^2 \otimes dx^1)$, so
$$
\langle dx^1\wedge dx^2,dx^1\wedge dx^2\rangle = \tfrac 1 2.
$$
More generally, the tensor inner product is $k!$ times the pointwise Hodge inner product on $k$-forms if you use the determinant convention, and $(1/k!)$ times if you use the Alt convension.
