Based on John Omielan's outline, I will prove below that every integer $a\geq 122$ has a representation with exponents $s_k\in\{1,2\}$.

Let us use the fact that every integer exceeding $33$ is a sum of distinct triangular numbers. Given this fact, we consider $a\geq 68$, and we define $c$ to be the largest nonnegative integer such that $$f(c):=\sum_{k=0}^c(k+2)$$ satisfies $$f(c)\equiv a\pmod{2}\qquad\text{and}\qquad f(c)\leq a-68.$$ Then $$f(c+2)=f(c)+(2c+7)>a-68,$$ that is, $$68\leq a-f(c)<2c+75.$$ By the quoted fact, there are distinct nonnegative integers $k_1,\dotsc,k_m$ such that $$a-f(c)=\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2).$$ We can ensure that no $k_\ell$ exceeds $c$ by imposing the condition $$(c+2)(c+3)\geq 2c+75,$$ which is certainly satisfied for integers $c>6$. Since we have already agreed on $f(c+2)>a-68$, it suffices to assume that $a-68\geq f(8)=54$, that is, $a\geq 122$.

To sum up, every integer $a\geq 122$ can be written as $$a=f(c)+\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2),$$ where the numbers $k_\ell\in\{0,\dotsc,c\}$ are distinct. Hence if we define $$s_k:=\begin{cases} 2,&k\in\{k_1,\dotsc,k_m\},\\ 1,&k\notin\{k_1,\dotsc,k_m\}, \end{cases}$$ then we obtain $$a=\sum_{k=0}^c(k+2)^{s_k}.$$

Based on John Omielan's outline, I will prove below that every integer $a\geq 122$ has a representation with exponents $s_k\in\{1,2\}$.

Let us use the fact that every integer exceeding $33$ is a sum of distinct triangular numbers. Given this fact, we consider $a\geq 68$, and we define $c$ to be the largest nonnegative integer such that $$f(c):=\sum_{k=0}^c(k+2)$$ satisfies $$f(c)\equiv a\pmod{2}\qquad\text{and}\qquad f(c)\leq a-68.$$ Then $$f(c+2)=f(c)+(2c+7)>a-68,$$ whence $a-f(c)$ is an even number from $\{68,\dotsc,2c+74\}$. By the quoted fact, there are distinct nonnegative integers $k_1,\dotsc,k_m$ such that $$a-f(c)=\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2).$$ We can ensure that no $k_\ell$ exceeds $c$ by imposing the condition $$(c+2)(c+3)>2c+74,$$ which is certainly satisfied for integers $c>6$. Since we have already agreed on $f(c+2)>a-68$, it suffices to assume that $a-68\geq f(8)=54$, that is, $a\geq 122$.

To sum up, every integer $a\geq 122$ can be written as $$a=f(c)+\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2),$$ where the numbers $k_\ell\in\{0,\dotsc,c\}$ are distinct. Hence if we define $$s_k:=\begin{cases} 2,&k\in\{k_1,\dotsc,k_m\},\\ 1,&k\notin\{k_1,\dotsc,k_m\}, \end{cases}$$ then we obtain $$a=\sum_{k=0}^c(k+2)^{s_k}.$$

I found John Omielan's outline hard to follow, so here is my own streamlined version.

We shall use the fact that every integer exceeding $33$ is a sum of distinct triangular numbers. Given this fact, we consider $a\geq 66$, and we define $c$ to be the largest nonnegative integer such that $$f(c):=\sum_{k=0}^c(k+2)$$ satisfies $$f(c)\equiv a\pmod{2}\qquad\text{and}\qquad f(c)\leq a-66.$$ Then $$f(c+2)=f(c)+(2c+7)>a-66,$$ that is, $$66\leq a-f(c)<2c+73.$$ By the quoted fact, there are distinct nonnegative integers $k_1,\dotsc,k_m$ such that $$a-f(c)=\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2).$$ We can ensure that no $k_\ell$ exceeds $c$ by imposing the condition $$(c+2)(c+3)\geq 2c+73,$$ which is certainly satisfied for integers $c>6$. Since we have already agreed on $f(c+2)>a-66$, it suffices to assume that $a-66\geq f(8)=54$, that is, $a\geq 120$.

To sum up, every integer $a\geq 120$ can be written as $$a=f(c)+\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2),$$ where the numbers $k_\ell\in\{0,\dotsc,c\}$ are distinct. Hence if we define $$s_k:=\begin{cases} 2,&k\in\{k_1,\dotsc,k_m\},\\ 1,&k\notin\{k_1,\dotsc,k_m\}, \end{cases}$$ then we obtain $$a=\sum_{k=0}^c(k+2)^{s_k}.$$

Based on John Omielan's outline, I will prove below that every integer $a\geq 122$ has a representation with exponents $s_k\in\{1,2\}$.

Let us use the fact that every integer exceeding $33$ is a sum of distinct triangular numbers. Given this fact, we consider $a\geq 68$, and we define $c$ to be the largest nonnegative integer such that $$f(c):=\sum_{k=0}^c(k+2)$$ satisfies $$f(c)\equiv a\pmod{2}\qquad\text{and}\qquad f(c)\leq a-68.$$ Then $$f(c+2)=f(c)+(2c+7)>a-68,$$ that is, $$68\leq a-f(c)<2c+75.$$ By the quoted fact, there are distinct nonnegative integers $k_1,\dotsc,k_m$ such that $$a-f(c)=\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2).$$ We can ensure that no $k_\ell$ exceeds $c$ by imposing the condition $$(c+2)(c+3)\geq 2c+75,$$ which is certainly satisfied for integers $c>6$. Since we have already agreed on $f(c+2)>a-68$, it suffices to assume that $a-68\geq f(8)=54$, that is, $a\geq 122$.

To sum up, every integer $a\geq 122$ can be written as $$a=f(c)+\sum_{\ell=1}^m (k_\ell+1)(k_\ell+2),$$ where the numbers $k_\ell\in\{0,\dotsc,c\}$ are distinct. Hence if we define $$s_k:=\begin{cases} 2,&k\in\{k_1,\dotsc,k_m\},\\ 1,&k\notin\{k_1,\dotsc,k_m\}, \end{cases}$$ then we obtain $$a=\sum_{k=0}^c(k+2)^{s_k}.$$