Let's go ahead and prove Eric Wofsey's claim (see his comment under the question), that if $H$ is an infinite-dimensional Hilbert space with an orthonormal basis $U$ of cardinality $\kappa$, then any Hamel basis $B$ of $H$ has cardinality $|B| = \kappa^{\aleph_0}$. Notice this implies that some (real or complex) vector spaces do not admit a Hilbert space structure; e.g., any whose dimension is the limit $\lambda$ of cardinals $\alpha_n$ defined by $\alpha_0 = \aleph_0$ and $\alpha_{n+1} = 2^{\alpha_n}$ (this $\lambda$ has countable cofinality and hence can't be of the form $\lambda = \kappa^{\aleph_0}$, else $\lambda = \lambda^{\aleph_0}$ which would contradict König's theorem, as mentioned in an earlier comment).
First remark: $|B| = |H|$. Since $H$ is infinite-dimensional, we have $|B| \geq 2^{\aleph_0} = |\mathbb{C}|$ (see for example the last two paragraphs of this M.SE post: https://math.stackexchange.com/a/547888/43208), and we have $|H| = |B|\cdot |\mathbb{C}|$ by the lemma at this MO post: https://mathoverflow.net/a/49572/2926). Combining these two, $|B| = |B| \cdot |B| \geq |B| \cdot |\mathbb{C}| = |H|$, and obviously $|B| \leq |H|$, so $|B| = |H|$.
Let $\langle -, -\rangle$ be the inner product. For each $h \in H$, let $N_h = \{e \in U: \langle h, e\rangle \neq 0\}$; notice this is at most countable. Consider $H_{fin} = \{h \in H: |N_h| < \infty\}$ and its complement, which I will denote as $H_\infty$.
Each $h \in H_\infty$ determines and is uniquely determined by the countably infinite set $N_h \subseteq U$ together with the corresponding element $\sum_{e \in N_h} \langle h, e\rangle e$ in $l^2(N_h)$. In this way we have a bijection
$$H_\infty \to \bigcup_{N \in X} L(N) \qquad (1)$$
where $X$ is the collection of countably infinite subsets $N$ of $U$ and $L(N)$ is the subset of $l^2(N)$ whose elements $h$ satisfy $\langle h, e\rangle \neq 0$ for all $e \in N$.
Lemma: If $A$ is an infinite set and $\alpha$ a cardinal such that $\alpha \leq |A|$, then the number of subsets of $A$ of size $\alpha$ is $|A|^\alpha$.
Proof: For each inclusion of a subset of size $\alpha$, we can choose an injective function $\alpha \to A$ with the same image, belonging to the set of all such functions which has cardinality $|A|^\alpha$, so $|A|^\alpha$ is an upper bound. On the other hand, a function $\alpha \to A$ can be identified with its graph, a subset of $\alpha \times A$ of size $\alpha$, and $|\alpha \times A| = |A|$ (excepting $\alpha = 0$ which is trivial), so $|A|^\alpha$ is also a lower bound. $\Box$
Applying this lemma to the bijection (1), we have $|X| = \kappa^{\aleph_0}$. Also we have $|L(N)| = 2^{\aleph_0}$ for each $N$, so $|H_\infty| = \kappa^{\aleph_0} \cdot 2^{\aleph_0} = \kappa^{\aleph_0}$.
Similarly, we have an injection
$$H_{fin} \to \bigcup_{n \geq 0} X_n \times \mathbb{C}^n \qquad (2)$$
where $X_n$ is the collection of subsets of $U$ of finite cardinality $n$. Since $|X_n| = \kappa^n = \kappa$ by the lemma, we have $|H_{fin}| \leq \aleph_0 \cdot \kappa \cdot 2^{\aleph_0} = \max\{\kappa, 2^{\aleph_0}\} \leq \kappa^{\aleph_0}$.
Finally, $\kappa^{\aleph_0} = |H_\infty| \leq |H| = |H_\infty| + |H_{fin}| \leq \kappa^{\aleph_0} + \kappa^{\aleph_0} = \kappa^{\aleph_0}$, so $|H| = \kappa^{\aleph_0}$ as claimed.
Added in response to Sushil's comment below: The converse, that if $\kappa = \kappa^{\aleph_0}$ then there exists a Hilbert space of algebraic dimension $\kappa$, is fairly immediate. Indeed, suppose $B$ is a set of cardinality $\kappa$, and let $V$ be the $\mathbb{C}$-vector space $V$ consisting of formal $\mathbb{C}$-linear combinations of elements of $B$. Assign to $V$ the unique inner product that makes $B$ an orthonormal set of $V$. Then the completion of $V$ with respect to the norm of the inner product is a Hilbert space $H$ whose algebraic dimension is $\kappa$. This is because $B$ is an orthonormal basis of $H$ (by construction of $H$), so the claim of the very first paragraph of this answer shows that any Hamel basis of $H$ has cardinality $\kappa^{\aleph_0} = \kappa$.
