If $a_1,a_2,\cdots,a_{25}$ denote the number of elements that are colored with color $i,$ then we have $a_1+a_2+\cdots+a_{25}=61,$ and then we seek to minimize $m=2^{a_1}+2^{a_2}+\cdots+2^{a_{25}}-25.$ We can use a smoothing argument: If $a+b$ is constant, then $2^a+2^b$ is minimized at $a=b$ from AM-GM, and similarly if $a+b$ is odd than we should let one exponent be $[(a+b)/2]$ and the other $[(a+b)/2]+1.$ Thus, we want the $a_i$ to be as close to $61/25=2.44$ as possible - so the exponents will be only two's and three's, and as many two's as possible. Then if there are $x$ 2's and $y$ 3's, we have $x+y=25, 2x+3y=65$ from which $x= 10, y= 15.$ $m$ is minimized at $10(2^2)+15(2^3)-25=135.$

If $x_1,x_2,\dots,x_{25}$ denote the number of elements that are colored with color $i$, then we have $x_1+x_2+\dots+x_{25}=61$. We want to minimize $$m=\sum_{i=1}^{25}(2^{x_i}-1)=2^{x_1}+2^{x_2}+\cdots+2^{x_{25}}-25.$$If there exist $i,j\in\{1,2,\dots,25\}$ such that $x_i-x_j\geq 2$ then $2^{x_i}+2^{x_j}>2^{x_i-1}+2^{x_j+1}$. We can reduce value of $m$ by replace $x_i,x_j$ with $x_{i}-1,x_{j}+1$. Thus, the minimum value of $m$ will occur when $|x_i-x_j|\leq1$ for all $i,j$. $$1\geq|x_i-x_1|\geq x_i-x_1\implies x_1+1\geq x_i\implies 25x_1+25\geq\sum_{i=1}^{25}x_i=61\implies x_1\geq \left\lceil\frac{36}{25}\right\rceil=2$$$$1\geq|x_1-x_i|\geq x_1-x_i\implies x_i\geq x_1-1\implies 61=\sum_{i=1}^{25}x_i\geq 25x_1-25\implies 3=\left\lfloor{\frac{86}{25}}\right\rfloor\geq x_1$$Similarly, $x_1,x_2,\dots,x_{25}\in\{2,3\}$. Let $a$ out of $25$ numbers, $x_1,x_2,\dots,x_{25}$ be $2$. $2(a)+3(25-a)=x_1+x_2+...+x_{25}=61\implies a=14, 25-a=11$ Minimum value of $m$ occur when $$\{x_1,x_2,...,x_{25}\}=\{\underbrace{2,2,\dots,2}_{14\text{ }2\text{'s}},\underbrace{3,3,...,3}_{11\text{ }3\text{'s}}\}.$$$\min(m)=14\cdot 2^2+11\cdot 2^3-25=119$. $\blacksquare$