For $p_1^2 + p_2^2$ (where $p_1$ and $p_2$ are two distinct primes) to be equal to $2^n$ (where $n \ge 2$), we must have both $p_1$ and $p_2$ odd. This would give $p_1^2 + p_2^2 \equiv 2 \pmod 4$, while $2^n \equiv 0 \pmod 4$, a contradiction.

For $p_1^2 + p_2^2 + p_3^2$ (where $p_1, p_2$ and $p_3$ are three distinct primes) to be equal to $2^n$ (where $n \ge 2$), we must have (say) $p_1 = 2$ and both $p_2$ and $p_3$ odd. This would give $p_1^2 + p_2^2 + p_3^2 \equiv 2 \pmod 4$, while $2^n \equiv 0 \pmod 4$, a contradiction.

For $p_1^2+p_2^2+p_3^2+p_4^2$ (where $p_1, p_2, p_3$ and $p_4$ are four distinct primes) to be equal to $2^n$ (where $n \ge 3$), we must have all four of these primes odd, say $p_i = 2m_i+1$, $i = 1, 2, 3, 4$. This would give $\sum_{i=1}^4 p_i^2 = \sum_{i=1}^4 (2m_i+1)^2 = \sum_{i=1}^4 (4m_i(m_i+1)+1) = 8L+4 = 2^n$ for some integer $L$, since each $m_i(m_i+1)$ is even. Then $2L+1 = 2^{n-2}$ would be a contradiction, as $2^{n-2}$ is even if $n \ge 3$.

We see that the smallest $k > 1$ that we are looking for must satisfy $k \ge 5$. Now, for the sum of the squares of five distinct primes to be a power of $2$, one of them must be $2$. The sum of the squares of the five smallest primes is not a power of two, but the very next choice of smallest primes does indeed satisfy the requirement: $2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 256 = 2^8$. Hence $k = 5$ is the smallest $k$ we are looking for.