$\def\supp{\mathop{\mathrm{supp}}}$ Surely, in the ``Almost equivalent question'' you assume that the $d$'s are square-free.

Denote $R=\mathbb Z[\sqrt{p_1},\dots,\sqrt{p_n}]$. We prove that in its group of units $U_R$, the elements $u_d=n_d+m_d\sqrt{d}$ (for all possible $d$) are independent (even modulo roots of unity $\pm1$), thus establishing a negative answer to both questions.

For that purpose, implement a proof of Dirichlet's unit theorem (see, e.g., p. 87 here). $R$ has $2^n$ automorphisms indexed by $I=(i_1,\dots,i_n)\in C=\{0,1\}^n$ and given by $\sigma_I(\sqrt{p_k})=(-1)^{i_k}\sqrt{p_k}$. Consider the group homomorphism $L\colon U_R\to\mathbb R^{C}$ given by $L(u)_I=|\ln \sigma_I(u)|$. It suffices to show that all the $v_d=L(u_d)$ are linearly independent. We show that they are linearly independent along with $v_0$ all whose coordinates are ones.

If $d=\prod_{j\in J}p_j$, then $L(u_d)_I$ attains a constant for all $I$ with $|J\cap \supp I|$ even, and another constant for all other $I$. So, modulo $v_0$, we may replace $v_d$ with $w_J$ such that $(w_J)_I=(-1)^{|J\cap \supp I|}$. All such $w_J$ (for all $J$) are well known to be independent (in fact, they consist just of the values of $\prod_{j\in J}(1-2x_j)$ on $C$).

$\def\supp{\mathop{\mathrm{supp}}}$ Surely, in the ``Almost equivalent question'' you assume that the $d$'s are square-free.

Denote $R=\mathbb Z[\sqrt{p_1},\dots,\sqrt{p_n}]$. We prove that in its group of units $U_R$, the elements $u_d=n_d+m_d\sqrt{d}$ (for all possible $d$) are independent (even modulo roots of unity $\pm1$), thus establishing a negative answer to both questions.

For that purpose, implement a proof of Dirichlet's unit theorem (see, e.g., p. 87 here). $R$ has $2^n$ automorphisms indexed by $I=(i_1,\dots,i_n)\in C=\{0,1\}^n$ and given by $\sigma_I(\sqrt{p_k})=(-1)^{i_k}\sqrt{p_k}$. Consider the group homomorphism $L\colon U_R\to\mathbb R^{C}$ given by $L(u)_I=|\ln \sigma_I(u)|$. It suffices to show that all the $v_d=L(u_d)$ are linearly independent. We show that they are linearly independent along with $v_0$ all whose coordinates are ones.

If $d=\prod_{j\in J}p_j$, then $L(u_d)_I$ attains a constant for all $I$ with $|J\cap \supp I|$ even, and another constant for all other $I$. So, modulo $v_0$, we may replace $v_d$ with $w_J$ such that $(w_J)_I=(-1)^{|J\cap \supp I|}$. All such $w_J$ (for all $J$) are well known to be independent (in fact, they consist just of the values of $\prod_{j\in J}(1-2x_j)$ on $C$).

NB. In fact, we have proved a more general fact. If, for any $\varnothing\neq J\subseteq \{1,\dots,n\}$ we choose a number $$ u_J=n_J+m_J\sqrt{\prod_{j\in J} p_j}, $$ where $n_J,m_J$ are nonzero rational numbers, and we also choose $u_\varnothing \in \mathbb Q\setminus\{0!\pm1\}$, then all products of the chosen numbers (with integer exponents) are distinct.

Surely, in the ``Almost equivalent question'' you assume that the $d$'s are square-free.

Denote $R=\mathbb Z[\sqrt{p_1},\dots,\sqrt{p_n}]$. We prove that in its group of units $U_R$, the elements $u_d=n_d+m_d\sqrt{d}$ (for all possible $d$) are independent (even modulo roots of unity $\pm1$), thus establishing a negative answer to both questions.

For that purpose, implement a proof of Dirichlet's unit theorem (see, e.g., p. 87 here). $R$ has $2^n$ automorphisms indexed by $I=(i_1,\dots,i_n)\in C=\{0,1\}^n$ and given by $\sigma_I(\sqrt{p_k})=(-1)^{i_k}\sqrt{p_k}$. Consider the group homomorphism $L\colon U_R\to\mathbb R^{C}$ given by $L(u)_I=|\ln \sigma_I(u)|$. It suffices to show that all the $v_d=L(u_d)$ are linearly independent. We show that they are linearly independent along with $v_0$ all whose coordinates are ones.

If $d=\prod_{j\in J}p_j$, then $L(u_d)_I$ attains a constant for all $I$ with $|I\cap J|$ even, and another constant for all other $I$. So, modulo $v_0$, we may replace $v_d$ with $w_J$ such that $(w_J)_I=(-1)^{|I\cap J|}$. All such $w_J$ (for all $J$) are well known to be independent (in fact, they are the values of the monomials $\prod_{j\in J}x_j$ on $\{-1,1\}^n$).

$\def\supp{\mathop{\mathrm{supp}}}$ Surely, in the ``Almost equivalent question'' you assume that the $d$'s are square-free.

Denote $R=\mathbb Z[\sqrt{p_1},\dots,\sqrt{p_n}]$. We prove that in its group of units $U_R$, the elements $u_d=n_d+m_d\sqrt{d}$ (for all possible $d$) are independent (even modulo roots of unity $\pm1$), thus establishing a negative answer to both questions.

For that purpose, implement a proof of Dirichlet's unit theorem (see, e.g., p. 87 here). $R$ has $2^n$ automorphisms indexed by $I=(i_1,\dots,i_n)\in C=\{0,1\}^n$ and given by $\sigma_I(\sqrt{p_k})=(-1)^{i_k}\sqrt{p_k}$. Consider the group homomorphism $L\colon U_R\to\mathbb R^{C}$ given by $L(u)_I=|\ln \sigma_I(u)|$. It suffices to show that all the $v_d=L(u_d)$ are linearly independent. We show that they are linearly independent along with $v_0$ all whose coordinates are ones.

If $d=\prod_{j\in J}p_j$, then $L(u_d)_I$ attains a constant for all $I$ with $|J\cap \supp I|$ even, and another constant for all other $I$. So, modulo $v_0$, we may replace $v_d$ with $w_J$ such that $(w_J)_I=(-1)^{|J\cap \supp I|}$. All such $w_J$ (for all $J$) are well known to be independent (in fact, they consist just of the values of $\prod_{j\in J}(1-2x_j)$ on $C$).