In 2012, T. Miyazaki and A. Togb{e} gave all of the solutions of the Diophantine equations $(2am-1)^x+(2m)^y=(2am+1)^z$ and $b^x+2^y=(b+2)^z$ in positive integers $x,y,z,$ $a>1$ and $bge 5$ odd. In this paper, we propose a similar problem (which we call the shuffle variant of a Diophantine equation of Miyazaki and Togb{e}). Here we first prove that the Diophantine equation $(2am+1)^x+(2m)^y=(2am-1)^z$ has only the solutions $(a, m, x, y, z)=(2, 1, 2, 1, 3)$ and $(2,1,1,2,2)$ in positive integers $a>1,m,x,y,z$. Then using this result, we show that the Diophantine equation $b^x+2^y=(b-2)^z$ has only the solutions $(b,x, y, z)=(5, 2, 1, 3)$ and $(5,1,2,2)$ in positive integers $x,y,z$ and $b$ odd.
Suppose that $n$ is a positive integer. In this paper, we show that the exponential Diophantine equation $$(n-1)^{x}+(n+2)^{y}=n^{z}, ngeq 2, xyz eq 0$$ has only the positive integer solutions $(n,x,y,z)=(3,2,1,2), (3,1,2,3)$. The main tools on the proofs are Bakers theory and Bilu-Hanrot-Voutiers result on primitive divisors of Lucas numbers.