The method suggested by your professor is sometimes called the universal side divisor criterion for showing that a ring in non-Euclidean. The basic idea is quite simple. To prove that a domain is non-Euclidean it suffices to show that it lacks some property possessed by all Euclidean domain. Here this property is that if $x$ is a nonunit of minimal Euclidean value, then, modulo $x$, every element has remainder $0$ or a unit (else the remainder is a nonunit having Euclidean value smaller than $x$, contra hypothesis).
Based on your hint, in your case it seems that the only units of $R$ are $\pm 1$, so every element $w$ has remainder $\,0\,$ or $\,\pm1,\,$ i.e. $x$ divides $w$ or $w\mp1$. It remains to show that there is some $w$ for which this fails, thus completing the proof that $R$ is not Euclidean.
If you google "universal side divisor" you will find some insightful expositions, including worked examples. For example see Keith Conrad's Remarks about Euclidean domains. See also my answer here which outlines a proof that $\rm\: \mathbb Z[w],\ w = (1 + \sqrt{-19})/2\ $ is a non-Euclidean PID - based on sketch by Hendrik W. Lenstra.
Here's what I think is a nice way to find a few PIDs that aren't Euclidean. It's not quite elementary, but if you know a bit about number fields I think it's a lot easier and nicer than the normal drudge.
Let $K=\mathbb{Q}(\sqrt{-d})$ for $d>3$ squarefree, with ring of integers $\mathcal{O}_K$. Then the only units in $\mathcal{O}_K$ are $\pm1$.
Suppose $\mathcal{O}_K$ is Euclidean with Euclidean function $\varphi$. Then take $x \in \mathcal{O}_K\setminus\{0,\pm1\}$ with $\varphi(x)$ minimal. By definition, any element of $\mathcal{O}_K$ can be written in the form $px+r$ where $\varphi(r) < \varphi(x)$, so it must be that $r \in \{0,\pm1\}$, i.e. $|\mathcal{O}_K/(x)|$ is $2$ or $3$. In other words $\mathcal{O}_K$ has a principal ideal of norm $2$ or $3$.
So now we know that if $K = \mathbb{Q}(\sqrt{-d})$ has class number one, where $d>3$ is squarefree*, $K$ is a non-Euclidean PID if there are no elements in $\mathcal{O}_K$ of norm $\pm2$ or $\pm3$. As $K$ is a PID (and degree $2$ over $\mathbb{Q}$), this is equivalent to saying that $2$ and $3$ are inert. To find some examples then:
If $d = 3\pmod{4}$, $\mathcal{O}_K = \mathbb{Z}[\frac{1+\sqrt{-d}}{2}]$, the minimal polynomial of $\frac{1+\sqrt{-d}}{2}$ over $\mathbb{Q}$ is $f_d(X)=X^2-X+\frac{1+d}{4}$. Applying Dedekind's criterion gives that $d$ works provided that $f_d(X)$ is irreducible $\pmod{2}$ and $\pmod{3}$. This then gives that $d = 19$ works (which is the usual example), but also shows that $d = 43,67$ or $163$ work as well (I think!).
*It can be shown that this implies $d \in \{1,2,3,7,11,19,43,67,163\}$
Best Answer
There is no simple rule that classifies when $\mathbf Z[\sqrt{d}]$ is a PID or Euclidean when $d$ is squarefree and positive, and there is no reason to expect a simple rule. (But note, as Gerry Myerson points out, that $\mathbf Z[\sqrt{d}]$ is the "wrong ring" to be thinking about when $d \equiv 1 \bmod 4$ since the full ring of algebraic integers of $\mathbf Q(\sqrt{d})$ in that case is the bigger ring $\mathbf Z[(1+\sqrt{d})/2]$. The ring $\mathbf Z[\sqrt{d}]$ is never a UFD for $d \equiv 1 \bmod 4$ for structural reasons.)
The list of norm-Euclidean real quadratic rings is known. See the Wikipedia page for quadratic integers, but note that it is about when the full ring of integers is norm-Euclidean. That is never $\mathbf Z[\sqrt{d}]$ when $d \equiv 1 \bmod 4$, so if you discard such $d$ then you are left with $d = 2, 3, 6, 7, 11, 19$. Note that being a Euclidean domain is a weaker property than being norm-Euclidean: maybe the ring is Euclidean with respect to some bizarre function, not its (absolute) norm function. An example of that is $\mathbf Z[\sqrt{14}]$, which is not norm-Euclidean but was proved to be Euclidean by Malcolm Harper.
If you want to accept the Generalized Riemann Hypothesis for zeta-functions of number fields, then PID = Euclidean for real quadratic rings of algebraic integers. For example, if $d$ is positive, squarefree, and not $1 \bmod 4$ then $\mathbf Z[\sqrt{d}]$ is Euclidean (not necessarily norm-Euclidean!) if and only if it is a PID. Neither of these properties is easy to characterize, but the properties are equivalent to each other if you accept GRH. This is a special case of a theorem of Weinberger in 1973: GRH for zeta-functions of all number fields implies the ring of integers $\mathcal O_K$ of a number field $K$ is Euclidean if it is a PID (has class number $1$) and the unit group $\mathcal O_K^\times$ is infinite. Weinberger's paper is On Euclidean rings of algebraic integers, pp. 321-332 in "Analytic number theory'' (Proc. Sympos. Pure Math., Vol. XXIV). Amer. Math. Soc., Providence (1973).