Here is a way to proceed without too much algebraic number theory. Of course, I assume you know that $O_K=\mathbb{Z}[\sqrt{10}]$ (see all the comments)
Of course, the unlerlying idea is that $I$ has norm $2$, and that there is no element in $O_K$ having norm $\pm 2$, but we don't need that.
In what follows, $N=N_{K/\mathbb{Q}}$.
Assume that $I=(\alpha)$ for some $\alpha\in O_K=\mathbb{Z}[\sqrt{10}]$. So $\alpha\mid 2$ and $\alpha\mid \sqrt{10}$ in $O_K$. Since norms are multiplicative, taking the norms yields $N(\alpha)\mid 4$ and $N(\alpha)\mid -10$ in $\mathbb{Z}$, so $N(\alpha)\mid 2$. It follows that $N(\alpha)=\pm 1,\pm 2$. The case $N(\alpha)=\pm 1$ implies that $\alpha$ is a unit, and that $I=O_K$. I let you check this is not the case.
So $N(\alpha)=\pm 2$. Writing $\alpha=a+b\sqrt{10}$, we have to solve $a^2-10 b^2=\pm 2$. In particular, $\pm 2$ is a square in $\mathbb{Z}/10\mathbb{Z}$. You can check this is not the case. Hence $I$ is not principal.
Side comment: if you want to show that $N(I)=2$, then you have several ways to proceed, depending what you know in algebraic number theory.
First method (very elaborate): since $2\in I$, you can try to factor $(2)$ as a product of prime ideals and see what happens. Since $O_K=\mathbb{Z}[\sqrt{10}]$, the factorisation a prime number $p$ is reflected by the factorisation of $X^2-10$ mod $p$. Here, $X^2-10=X^2$ mod $p$, and then $(2)=(2,\sqrt{10})^2=I^2$ by a famous theorem of Dedekind.
In particular, $N(I^2)=N(I)^2= N(2O_K)=\vert N(2)\vert =4$, so $N(I)=2$.
Second method(elementary): Well, you can check that $I=\{a+b\sqrt{10}, a\in 2\mathbb{Z},b\in\mathbb{Z}\}$. This is not totally obvious, and does not comes directly from the definition of $I$, since a generic element of $I$ has the form $2z_1+\sqrt{10}z_2,$ with $z_1,z_2\in O_K $(and not $z_1,z_2\in\mathbb{Z}$), so this requires a bit of easy computation. Hence, a $\mathbb{Z}$-basis of $O_K$ is $(1,\sqrt{10})$, while a basis of $I$ is $(2,\sqrt{10})$. Thus, as an abelian group, we get $O_K/I\simeq \mathbb{Z}/2\mathbb{Z}$ and $N(I)=\vert O_K/I\vert =2$.
Best Answer
Yes, that is true. Use that ideals factor uniquely into prime ideals in Dedekind domains. We get $I^2 = \mathfrak{p}_1^{2e_1} \dots \mathfrak{p}_k^{2e_k}$ for $e_1, \dots, e_k \in \mathbb{N}$. Moreover, say $(\sqrt{d}) = \mathfrak{q}_1^{d_1} \dots q_{\ell}^{d_{\ell}}$. So we get $$ \mathfrak{q}_1^{2d_1} \dots \mathfrak{q}_{\ell}^{2d_{\ell}} = (\sqrt{d})^2 = (d) = I^2 = \mathfrak{p}_1^{2e_1} \dots \mathfrak{p}_k^{2e_k}.$$ Comparing primes yields $I = (\sqrt{d})$.