**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 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 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: 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 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 $X_{i,y}:=\min(y,X_i)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a supermartingale difference sequence and $Var\,X_{i,y}\le Var\,X_i$. So, 
\begin{align*}
	P(S_n\ge x)&\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^nX_{i,y}\ge x\Big) \\ 
	&\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^n[X_{i,y}-E(X_{i,y}|\mathcal F_{i-1})]\ge x\Big) \\ 
	&\le \sum_{i=1}^n P(X_i>y)+\exp\frac{-cx^2}{B^2+xy}
\end{align*}
by the highlighted statement, for all real $x>0$. So, 
\begin{equation*}
		P(|S_n|\ge x)\le\sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy}. 
\end{equation*}
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. 
$\Box$