= Solution
A <Riemannian metric> is a smooth positive-definite symmetric <inner product> $g_x$ on every <tangent space>. A <Levi-Civita connection> is a connection on the <tangent bundle> which is torsion-free, $\nabla_XY-\nabla_YX=[X,Y]$, and metric-compatible,
$$
Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ).
$$
Apply this identity with the cyclic triples $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and replace differences of covariant derivatives by <Lie brackets>. The result is the <Koszul formula>
$$
\boxed{\begin{aligned}
2g(\nabla_XY,Z)={}&Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)\\
&-g(X,[Y,Z])+g(Y,[Z,X])+g(Z,[X,Y]).
\end{aligned}}
$$
Its right side depends only on $g$ and the <vector fields>; nondegeneracy of $g$ determines $\nabla_XY$ uniquely. This proves uniqueness.
For existence, in a coordinate chart define
$$
\Gamma^k_{ij}=\frac12g^{k\ell}(\partial_i g_{j\ell}+\partial_jg_{i\ell}-\partial_\ell g_{ij}),\qquad
(\nabla_XY)^k=X^i\partial_iY^k+\Gamma^k_{ij}X^iY^j.
$$
This is linear over smooth functions in $X$ and obeys the Leibniz rule in $Y$. Symmetry $\Gamma^k_{ij}=\Gamma^k_{ji}$ proves zero torsion, and direct substitution gives
$$
\partial_i g_{jk}=g_{\ell k}\Gamma^\ell_{ij}+g_{j\ell}\Gamma^\ell_{ik},
$$
which proves <metric compatibility>. On chart overlaps the two resulting connections satisfy the same intrinsic Koszul identity and therefore agree by uniqueness. They glue globally. Thus the <existence and uniqueness of the Levi-Civita connection> holds on every <Riemannian manifold>.
Fix the curvature convention
$$
R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.
$$
For coordinate vectors $e_i$, define the <Riemann curvature tensor> components by $R_{ij,kl}=g(R(e_i,e_j)e_l,e_k)$. In these indices the Ricci and scalar contractions are
$$
\boxed{\operatorname{Ric}_{jl}=g^{ik}R_{ij,kl},\qquad
s=g^{jl}\operatorname{Ric}_{jl}=g^{ik}g^{jl}R_{ij,kl}}.
$$
Explicitly, $R_{ij,kl}=g_{ka}R^a{}_{l ij}$, where
$$
R^a{}_{l ij}=\partial_i\Gamma^a_{jl}-\partial_j\Gamma^a_{il}
+\Gamma^a_{ip}\Gamma^p_{jl}-\Gamma^a_{jp}\Gamma^p_{il}.
$$
These expressions define <Ricci curvature> and <scalar curvature> without ambiguity about which indices are contracted. With this convention a round unit $m$-sphere has $\operatorname{Ric}=(m-1)g$ and $s=m(m-1)$.
If $\operatorname{Ric}=\lambda g$ with a constant real $\lambda$, tracing gives $\boxed{s=m\lambda}$, so every <Einstein metric> has constant <scalar curvature>. The converse is not valid in arbitrary dimension. On $S^2\times S^1$ with the <product Riemannian metric> of the unit round sphere and a circle, the connection splits between factors. The mixed curvatures and circle curvature vanish, while the sphere has curvature one. Hence
$$
\boxed{\operatorname{Ric}=g_{S^2}\oplus0,\qquad s=2\text{ everywhere}}.
$$
A scalar multiple of the whole metric would require $\lambda=1$ on the sphere directions and $\lambda=0$ on the circle direction, which is impossible. This proves that <constant scalar curvature does not imply an Einstein metric>, and supplies a counterexample to the unrestricted requested equivalence.
In dimension two, the curvature tensor is determined by the <Gaussian curvature> $K$:
$$
R_{ij,kl}=K(g_{ik}g_{jl}-g_{il}g_{jk}),\qquad
\operatorname{Ric}=Kg,\qquad s=2K.
$$
There the valid equivalence is
$$
\boxed{s\text{ constant}\iff\operatorname{Ric}=\lambda g\text{ with constant }\lambda=s/2}.
$$
In dimension one both curvature and <scalar curvature> vanish identically. Without a two-dimensional restriction or another added geometric hypothesis, the general converse cannot be proved.
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