Let $P_1,\dots,P_k$ be polynomials over $\mathbf{C}$, no two of them being proportional.
Does there exist an integer $N$ such that $P_1^N,\dots,P_k^N$ are linearly independent?
Polynomials – Are Large Powers Linearly Independent?
ac.commutative-algebralinear algebrapolynomialsra.rings-and-algebras
Related Solutions
Here is a linearly independent subset of $\mathbb{R}$ with size $2^{\aleph_0}$.
Let $q_0, q_1, \ldots$ be an enumeration of $\mathbb{Q}$. For every real number $r$, let $$T_r = \sum_{q_n < r} \frac{1}{n!}$$ The proof that these numbers are linearly independent is similar to the usual proof that $e$ is irrational. (It's a cute problem; there's spoiler below.)
I think a similar trick might work for algebraic independence, but I don't recall having seen such a construction. Actually, John von Neumann showed that the numbers $$A_r = \sum_{n=0}^\infty \frac{2^{2^{[nr]}}}{2^{2^{n^2}}}$$ are algebraically independent for $r > 0$. [Ein System algebraisch unabhängiger zahlen, Math. Ann. 99 (1928), no. 1, 134–141.] A more general result due to Jan Mycielski seems to go through in ZF + DC perhaps just ZF in some cases. [Independent sets in topological algebras, Fund. Math. 55 (1964), 139–147.]
As for subspaces and subfields isomorphic to $\mathbb{R}$, the answer is no. (Since I'm not allowed to post any logic here, I'll refer you to this answer and let you figure it out.)
Well, I'll bend the rules a little... Consider a $\mathbb{Q}$-linear isomorphism $h:\mathbb{R}\to H$, where $H$ is a $\mathbb{Q}$-linear subspace of $\mathbb{R}$ (i.e. $h$ is an additive group isomorphism onto the divisible subgroup $H$ of $\mathbb{R}$). If $h$ Baire measurable then it must be continuous by an ancient theorem of Banach and Pettis. It follows that $h(x) = xh(1)$ for all $x \in \mathbb{R}$ and therefore $H = \mathbb{R}$. Shelah has produced a model of ZF + DC where all sets of reals have the Baire property, so any such $h$ in this model must be Baire measurable. A similar argument works if Baire measurable is replaced by Lebesgue measurable, but Solovay's model of ZF + DC where all sets of reals are Lebesgue measurable uses the existence of an inaccessible cardinal, and this hypothesis was shown necessary by Shelah.
Spoiler
Suppose for the sake of contradiction that $r_1 > r_2 > \cdots > r_k$ and $a_1,a_2,\ldots,a_k \in \mathbb{Z}$ are such that $a_1T_{r_1} + a_2T_{r_2} + \cdots + a_kT_{r_k} = 0$. Choose a very large $n$ such that $r_1 > q_n > r_2$. If $n$ is large enough that $$(|a_1| + |a_2| + \cdots + |a_k|) \sum_{m=n+1}^\infty \frac{n!}{m!} < 1$$ then the tail terms of $n!(a_1T_{r_1}+\cdots+a_kT_{r_k}) = 0$ must cancel out, and we're left with $$a_1 = -\sum_{m=0}^{n-1} \sum_{q_m < r_i} a_i \frac{n!}{m!} \equiv 0 \pmod{n}$$ If moreover $n > |a_1|$, this means that $a_1 = 0$. Repeat to conclude that $a_1 = a_2 = \cdots a_k = 0$.
Best Answer
The answer is yes. In fact, an even stronger claim is true: there exists some $N$ such that for all $n \geq N, \ P_{1}^{n}, \dots, P_{k}^n$ are linearly independent over $\mathbb{C}$.
For this we will use a generalization of the Mason-Stother's theorem which appears on the Wikipedia page (though I have taken the special case of the curve $C = \mathbb{P}^{1} (\mathbb{C})$ and written it in slightly different language.):
Let $q_1, \dots, q_{k}$ be polynomials such that $q_1 + \cdots + q_{k} = 0$ and every proper subset of $q_1, \dots, q_{k}$ is linearly independent. Then, $$\max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right)$$
Now, we can prove the claim by induction on $k$. For $k = 2$ it is obvious. Now, by induction for all $n$ large enough every proper subset of $P_{1}^{n}, \dots, P_{k}^{n}$ is linearly independent. Suppose for contradiction that there exist constants $\lambda_{1}, \dots, \lambda_{k}$ such that $$\lambda_1 P_{1}^n + \cdots + \lambda_{k} P_{k}^n = 0$$ Letting $q_i = \lambda_i P_{i}^{n}$, notice that $q_1, \dots, q_k$ satisfy the requirements of the lemma (we have assumed that $\lambda_i \neq 0$), and therefore $$n \leq \max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right) = \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( P_1 \cdots P_k \right) \right) - 1 \right)$$ but the right hand side is constant, and so for $n$ large we get a contradiction.