Suppose that $S_c:=\{\sqrt{n^2+a^2}:n\in\mathbb{N}\}$ is linearly dependent over $\mathbb{Q}$. This means that there is a finite list of rational numbers $a_1,\ldots,a_r$ so that $$ \sum_{n=1}^r a_n\sqrt{n^2+c^2} = 0.$$ Hence the set of $c$ values such that $S_c$ is $\mathbb{Q}$-linearly dependent is smaller than the set of finite sequences of rational numbers. Since the latter set is countable, so is the set of such $c$. Hence the set $S_c$ is $\mathbb{Q}$-linearly independent for almost all $c\in\mathbb{R}$. Of course, this Cantor argument is useless for proving anything about any particular $c$, such as $c=\pi$.

Suppose that $S_c:=\{\sqrt{n^2+c^2}:n\in\mathbb{N}\}$ is linearly dependent over $\mathbb{Q}$. This means that there is a finite list of rational numbers $a_1,\ldots,a_r$ so that $$ \sum_{n=1}^r a_n\sqrt{n^2+c^2} = 0.$$ Hence the set of $c$ values such that $S_c$ is $\mathbb{Q}$-linearly dependent is smaller than the set of finite sequences of rational numbers. Since the latter set is countable, so is the set of such $c$. Hence the set $S_c$ is $\mathbb{Q}$-linearly independent for almost all $c\in\mathbb{R}$. Of course, this Cantor argument is useless for proving anything about any particular $c$, such as $c=\pi$.