Your exact question has already been resolved and is today a simple matter of looking it up in the OEIS and sifting out the numbers of the form $4k + 1$.

$$d = 2, 3, 6, 7, 11 \textrm{ or } 19$$

But it was a long, slow process throughout the 20th century to arrive at this answer, as Ian Stewart and David Tall explain in their excellent book *Algebraic Number Theory and Fermat's Last Theorem*.

Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a, b \in \mathbb{Z}\}$, not the ring of integers of $\mathbb{Q}[\sqrt{d}]$.

Because you said this, it's necessary to sift out the numbers of the form $4k + 1$. Stewart & Tall (and many other authors in other books) show that if a domain is Euclidean then it is a principal ideal domain and a unique factorization domain (the converse doesn't always hold, but that's another story).

So if $d > 5$, $d \equiv 1 \pmod 4$ and $$m = \frac{d - 1}{4},$$ (that's an integer) then $d - 1 = 2^2 m = (-1)(1 - \sqrt d)(1 + \sqrt d)$ represents two distinct factorizations of the same number, which means that $\mathbb Z[\sqrt d]$ is not a unique factorization domain, which in turns means it can't be Euclidean and certainly not norm-Euclidean.

If you had asked about Euclidean in general, you question would be significantly more difficult. The Euclidean function for $\mathcal O_{\mathbb Q(\sqrt{69})}$ was only recently (just a couple of decades ago) discovered: the norm function requires two adjustments and is somewhat frustrating for practical applications.

And it has been proven that $\mathbb Z[\sqrt{14}]$ is Euclidean but not norm-Euclidean, but no one on Math.SE seems to know what the Euclidean function might be (it has been asked).