Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.
Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)
This "proof" of "Theorem 1" is not so non-serious as it may look.
Perhaps more seriously, we have
Theorem 2: The following statement is false:
There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that
\begin{equation*}
X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0}
\end{equation*}
we have
\begin{equation*}
P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1}
\end{equation*}
for all real $x>0$.
Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that, at least in the case when the $X_i$'s are conditionally symmetric (given $\mathcal F_{i-1}$), that theorem implies the Rosenthal-type inequality
\begin{equation*}
ES_n^4\ll B^4+A^{(4)}_n,
\end{equation*}
where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and
\begin{equation*}
A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p.
\end{equation*}
Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:
Lemma 1: If the highlighted statement is true, then for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have
\begin{equation*}
ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2}
\end{equation*}
This lemma will be proved at the end of this answer.
Now consider the following construction of a conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let
\begin{equation*}
V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}},
\end{equation*}
where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$, and for natural $k\ge2$ let
\begin{equation*}
X_k:=S_{k-1}V_k,
\end{equation*}
where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$,
\begin{equation*}
S_k=S_{k-1}(1+V_k).
\end{equation*}
So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence
\begin{equation*}
M_k^{(p)}=\prod_{j=2}^k E(1+V_j)^p.
\end{equation*}
In particular,
\begin{equation*}
M_k^{(2)}=\prod_{j=2}^k (1+a_k^2)=\prod_{j=2}^k \Big(1+\frac1{k\ln k}\Big)
=\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\}
=(\ln k)^{1+o(1)}
\end{equation*}
(as $k\to\infty$). Similarly,
\begin{equation*}
M_k^{(4)}=\prod_{j=2}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)},
\end{equation*}
\begin{equation*}
M_k^{(6)}=\prod_{j=2}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}.
\end{equation*}
Hence,
\begin{equation*}
A^{(6)}_n=1+\sum_{k=2}^n M_{k-1}^{(6)}a_k^6\ll1+\sum_{k=2}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1.
\end{equation*}
Also, we may take
\begin{equation*}
B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}.
\end{equation*}
So, for $n\to\infty$ the right-hand side of \eqref{2} is
\begin{equation*}
B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)},
\end{equation*}
whereas the left-hand side of \eqref{2} is
\begin{equation*}
ES_n^4=M_n^{(4)}=(\ln n)^{6+o(1)}.
\end{equation*}
Thus, \eqref{2} fails to hold for large enough $n$.
It remains to give
Proof of Lemma 1: Suppose the highlighted statement is true. Take any conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=X_i\,1(|X_i|\le y)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a martingale difference sequence with $|X_{i,y}|\le y$ and $Var\,X_{i,y}\le Var\,X_i$ for all $i$. So,
\begin{align*}
P(|S_n|\ge x)&\le\sum_{i=1}^n P(|X_i|>y)+P\Big(\Big|\sum_{i=1}^nX_{i,y}\Big|\ge x\Big) \\
&\le \sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy}
\end{align*}
by the highlighted statement, for all real $x>0$.
Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have
\begin{align*}
ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\
&\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\
& +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\
&\ll A^{(6)}_n/B^2+B^4.
\end{align*}
This completes the proof of Lemma 1 and thus the proof of Theorem 2.
$\Box$
Whereas, as has just been shown, the highlighted statement is false even for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$, note Theorem 3.6, which implies that for any conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n X_i^2\le B^2$ for some real $B>0$, we have
\begin{equation*}
P\Big(\Big|\sum_{i=1}^n X_i\Big|\ge x\Big)\le2\exp\frac{-x^2}{2B^2}
\end{equation*}
for all real $x>0$.
