I think the answer is no. I found the following counterexample.

Let $K$ be the field of complex Hahn series with real exponents, i.e.
$$K = \left\{ f = \sum_{r \in \mathbb{R}} a_r X^r ; \, \operatorname{supp}(f) \textrm{ is a well-ordered subset of } (\mathbb{R}, \ge) \right\},$$
where $\operatorname{supp}( \sum_{r \in \mathbb{R}} a_r X^r ) := \{r \in \mathbb{R} ; \, a_r \neq 0\}$.

Let $R$ be the subring of $K$, defined as $R = \{f \in K ; \, \operatorname{supp}(f) \subseteq \mathbb{R}_{\ge 0}\}$. This is a local ring, with maximal ideal $\mathfrak{m} = \{f \in K ; \operatorname{supp}{f} \subseteq \mathbb{R}_{>0}\}$. The residue field $k= R/m \cong \mathbb{C}$ is an $R$-module.

I claim that $M = R/\mathfrak{m}$ is a counterexample, i.e. $\operatorname{Ext}_R^1(M,R/I) = 0$ for all ideals $I$ of $R$, yet $M$ is not a projective $R$-module.

Proof:

$M$ is not projective: If $M$ were projective, then $0 \to \mathfrak{m} \to R \to M \to 0$ would split, hence $\mathfrak{m}$ would be a quotient of $R$, so in particular it could be generated by one element. However $\mathfrak{m}$ is not a finitely generated ideal.

$\operatorname{Ext}_R^1(M,R/I) = 0$:
The short exact sequence $0 \to \mathfrak{m} \to R \to M \to 0$ gives the long exact sequence
$$
\begin{split}
0 &\to \operatorname{Hom}(M, R/I) \to \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I) \to \\
 &\to \operatorname{Ext}^1(M, R/I) \to \operatorname{Ext}^1(R,R/I) \to \dotsm
\end{split}
$$

Here $\operatorname{Ext}^1(R,R/I) = 0$ since $R$ is projective. So it is enough to prove that $R/I = \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I)$ is surjective.

The ideals of $R$ are easy to describe: If $c \in \mathbb{R}_{\ge 0}$, then let $I_{\ge c} = (X^c)$ and $I_{>c} = (X^r ; \, r > c)$. Then every nonzero ideal is of the form $I_{\ge c}$ or $I_{>c}$. In particular $\mathfrak{m} = I_{>0}$.

If $I=0$:
we need that $R = \operatorname{Hom}(R,R) \to \operatorname{Hom}(\mathfrak{m}, R)$ is surjective (in fact it is bijective).
This is not hard to check (for $\varphi \in \operatorname{Hom}(\mathfrak{m}, R)$,
show that $\varphi(X^r) \in I_{\ge r}$, and $X^{-r} \varphi(X^r) \in R$ is independent of $r>0$).

If $I=I_{\ge c}$:
Let $\varphi \colon \mathfrak{m} \to R/I_{\ge c}$ be a homomorphism.
We need to show that there is an $h \in R$ such that $\varphi(f) = f h + I_{\ge c}$.
To prove this, look at $\varphi(X^r)$ and take $r \to 0$.
If $r<c$, then $\varphi(X^r)$ will determine $h$ modulo $I_{\ge c-r}$.
To finish the proof, we need that if $E \subseteq [0, c)$ such that $E \cap [0,d)$ is well-ordered for all $d<c$, then $E$ is also well-ordered.

For $I=I_{> c}$, the proof is almost the same.