Let $k = 2$. If $n_1 = 5^l$ and $n_2 = 2 \cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+1}$. If $n_1 = 3 \cdot 5^l$ and $n_2 = 4 \cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+2}$. In both cases, $l$ is an arbitrary non-negative integer. Hence, the equation $n_1^2 + n_2^2 = 5^l$ has solutions for all positive integers $t$.

Next, we consider the case $k = 3$. Let $n_1 = 3a_1$, $n_2 = 4a_1$ and $n_3 = 5^2 a_2$. Then $n_1^2 + n_2^2 + n_3^2 = 5^2(a_1^2 + a_2^2)$. For every non-negative integer $m$ there exists a solution to the equation $a_1^2 + a_2^2 = 5^{1+m}$. so we have shown that the equation $n_1^2 + n_2^2 + n_3^2 = 5^{3+m}$ has solutions for all non-negative integers $m$.

For $k \ge 3$ we proceed by induction. Assume that the equation $a_1^2 + a_2^2 + \dots + a_l^2 = 5^{l+m}$ has positive integer solutions for all non-negative integers $m$ and for all positive integers $l$ such that $2 \le l \le k$. We wish to show that in this case the equation $n_1^2 + n_2^2 + \dots + n_k^2 + n_{k+1}^2 = 5^{k+1+m}$ also has positive integer solutions for all non-negative integers $m$.

Since $5^{k+1+m} = 5^{k+m} + 4 \cdot 5^{k+m}$ and $k > k-1 \ge 2$, the induction hypothesis implies that for all non-negative integers $i$ there exist positive integers $a_1, a_2, \dots, a_{k-1}$, such that
$$a_1^2 + a_2^2 + \dots + a_{k-1}^2 = 5^{k-1+i}$$
In particular, this is true for $i = m + 1$. so there exist positive integers $a_1, a_2, \dots, a_{k-1}$, such that
$$a_1^2 + a_2^2 + \dots + a_{k-1}^2 = 5^{k+m}$$

Since $3^2 + 4^2 = 5^2$ and $5^2 + 10^2 = 5^3$, such numbers $n_1$ and $n_2$ exist for $k=2$, $m=0$ and for $k=2$, $m=1$. Using induction on $m$ we show that the solutions exist for any $m$. Let $m \ge 3$. Assume that there exist positive integers $a_1, a_2$, such that $a_1^2 + a_2^2 = 5^m$. Then
$$5^{m+2} = 5^2 \cdot 5^m = 5^2(a_1^2 + a_2^2) = (5a_1)^2 + (5a_2)^2.$$
so $5^{m+2}$ is the sum of the squares of two positive integers. We have considered the cases $m=0$ and $m=1$ separately, so by induction the claim for $k=2$ follows for all non-negative integers $m$.

Since $5^3 = 3^2 + 4^2 + 10^2$ and $5^4 = 9^2 + 12^2 + 20^2$, we can use a similar argument for $k=3$ and show that the equation $n_1^2 + n_2^2 + n_3^2 = 5^{3+m}$ also has positive integer solutions for all non-negative integers $m$.

Let $k \ge 3$. We will use induction on $k$. Assume that for all non-negative integers $m$ there exist positive integers $a_1, a_2, \dots, a_k$, such that
$$a_1^2 + a_2^2 + \dots + a_k^2 = 5^{k+m}$$
and, since $k-1 \ge 2$, positive integers $b_1, b_2, \dots, b_{k-1}$, such that
$$b_1^2 + b_2^2 + \dots + b_{k-1}^2 = 5^{k-1+m}$$
If $k+m$ is even, then $5^{k+m+1} = 5^{k+m} + 4 \cdot 5^{k+m} = a_1^2 + a_2^2 + \dots + a_k^2 + 4 \cdot 5^{k+m} = a_1^2 + a_2^2 + \dots + a_k^2 + (2 \cdot 5^{\frac{k+m}{2}})^2$. So $5^{k+m+1}$ is a sum of $k+1$ squares of positive integers.

Let $k+m$ be odd. Since $k+m-3 \ge 0$, we can find two positive integers $c_1, c_2$, such that $c_1^2 + c_2^2 = 5^{2+(k+m-3)}$. Then
$$\begin{aligned} 5^{k+m+1} &= 5^2 \cdot 5^{k+m-1} = (9+16)5^{k+m-1} \\ &= 9 \cdot 5^{k-1+m} + 16 \cdot 5^{2+k+m-3} \\ &= 9(b_1^2 + b_2^2 + \dots + b_{k-1}^2) + 16(c_1^2 + c_2^2) \\ &= (3b_1)^2 + (3b_2)^2 + \dots + (3b_{k-1})^2 + (4c_1)^2 + (4c_2)^2. \end{aligned}$$

Since $m$ was an arbitrary non-negative integer, this concludes the induction.