A solution indeed exists! Or does it?

We may assume $y=2k$ . Then

$615=3\times 5\times 41=2^{2k}-x^2=(2^k-x)(2^k+x)$ . Then we need to put factors of $615$ in two groups $A$ and $B$ .

$A \times B= (2^k-x)(2^k+x)$ . if $A,B$ are known, we can solve a system of linear equations

$\begin{cases} 2^k-x=A \\ 2^k+x=B \end{cases}$

if the equations are added, we get $2^{k+1}=A+B$ . So, we need to group factors of $615$ , such that their addition would be a power of two. there are three possibilities and only one of them work ( $A=5,B=3\times 41$ ). Therefore, $A+B=2^{k+1}=128 \implies k=6 \implies 2k=y=12$