There is a simple criterion. I learned from Jacob Lurie's paper, Lemma 5.5. His argument works, but since his statement does not cover the current situation, let me phrase an adapted statement and transcribe his proof, hopefully correct.

Let $f\colon A\to B$ be a map of associative rings. Let $A''$ denote the cokernel of $f$ as an abelian group, which carries a canonical $A$-$A$-bimodule structure. Then the map $f$ exhibits $B$ as a faithfully flat left $A$-module if and only if

1. the map $f$ is injective, and
2. the left $A$-module $A''$ is flat.

Proof. We start with assuming the injectivity of $f$ and the flatness of $A''$. Then the left $A$-module $B$ is an extension of $A''$ by $A$. Since flat modules are stable under extensions, the left $A$-module $B$ is flat. For every right $A$-module $N$, we have an exact sequence $$\DeclareMathOperator\Tor{Tor} 0=\Tor^A_1(N,A'')\to N\to N\otimes_AB\to N\otimes_AA''\to0. $$ If $N\otimes_AB=0$, then $N=0$. It follows that the left $A$-module $B$ is faithfully flat.

Now we assume that the map $f$ exhibits $B$ as a faithfully flat left $A$-module. Then the map $B=A\otimes_AB\xrightarrow{f\otimes1}B\otimes_AB$ of abelian groups is injective, since it admits a retract given by the multiplication map $B\otimes_AB\to B$. Since the left $A$-module $B$ is faithfully flat, it follows that the map $f$ of right $A$-modules is injective.

It remains to see that the left $A$-module $A''$ is flat. Let $N'\to N$ be any injective map of right $A$-modules. We have to see that the map $N'\otimes_AA''\to N\otimes_AA''$ of right $A$-modules is injective, which is equivalent to the injectivity of $N'\otimes_AA''\otimes_AB\to N\otimes_AA''\otimes_AB$ by faithful flatness of the left $A$-module $B$. We now examine the short exact sequence $$ 0\to B=A\otimes_AB\xrightarrow{f\otimes1}B\otimes_AB\to A''\otimes_AB\to0 $$ of left $A$-modules (by left flatness of $B$). As previously, the map $f\otimes1$ of left $A$-module admits a retract given by the multiplication $B\otimes_AB\to B$ which is also left $A$-linear. It follows that the left $A$-module $A''\otimes_AB$ is a direct summand of the left $A$-module $B\otimes_AB$. Since the left $A$-module $B\otimes_AB$ is flat, so is the left $A$-module $A''\otimes_AB$, and the result follows.

There is a simple criterion. I learned from Jacob Lurie's paper Tannaka Duality for Geometric Stacks, Lemma 5.5. His argument works, but since his statement does not cover the current situation, let me phrase an adapted statement and transcribe his proof, hopefully correct.

Let $f\colon A\to B$ be a map of associative rings. Let $A''$ denote the cokernel of $f$ as an abelian group, which carries a canonical $A$-$A$-bimodule structure. Then the map $f$ exhibits $B$ as a faithfully flat left $A$-module if and only if

1. the map $f$ is injective, and
2. the left $A$-module $A''$ is flat.

Proof. We start with assuming the injectivity of $f$ and the flatness of $A''$. Then the left $A$-module $B$ is an extension of $A''$ by $A$. Since flat modules are stable under extensions, the left $A$-module $B$ is flat. For every right $A$-module $N$, we have an exact sequence $$\DeclareMathOperator\Tor{Tor} 0=\Tor^A_1(N,A'')\to N\to N\otimes_AB\to N\otimes_AA''\to0. $$ If $N\otimes_AB=0$, then $N=0$. It follows that the left $A$-module $B$ is faithfully flat.

Now we assume that the map $f$ exhibits $B$ as a faithfully flat left $A$-module. Then the map $B=A\otimes_AB\xrightarrow{f\otimes1}B\otimes_AB$ of abelian groups is injective, since it admits a retract given by the multiplication map $B\otimes_AB\to B$. Since the left $A$-module $B$ is faithfully flat, it follows that the map $f$ of right $A$-modules is injective.

It remains to see that the left $A$-module $A''$ is flat. Let $N'\to N$ be any injective map of right $A$-modules. We have to see that the map $N'\otimes_AA''\to N\otimes_AA''$ of right $A$-modules is injective, which is equivalent to the injectivity of $N'\otimes_AA''\otimes_AB\to N\otimes_AA''\otimes_AB$ by faithful flatness of the left $A$-module $B$. We now examine the short exact sequence $$ 0\to B=A\otimes_AB\xrightarrow{f\otimes1}B\otimes_AB\to A''\otimes_AB\to0 $$ of left $A$-modules (by left flatness of $B$). As previously, the map $f\otimes1$ of left $A$-module admits a retract given by the multiplication $B\otimes_AB\to B$ which is also left $A$-linear. It follows that the left $A$-module $A''\otimes_AB$ is a direct summand of the left $A$-module $B\otimes_AB$. Since the left $A$-module $B\otimes_AB$ is flat, so is the left $A$-module $A''\otimes_AB$, and the result follows.