You have the inductive case of $n = m$ giving
$$\sum_{k=1}^{m} \frac{1}{2^{k^2}} \leq 1-\frac{1}{2^{m^2}} \tag{1}\label{eq1A}$$
A common method to use induction with inequalities is to go from the inductive case, given above, to the case you're trying to prove, i.e., for $n = m + 1$, by changing the values so one side is true and then showing the other side is still true as well. In this case, the change is simplest to deal with on the left side as it just involves adding $\dfrac{1}{2^{(m+1)^2}}$. Thus, adding this value to both sides gives
$$\begin{equation}\begin{aligned}
\sum_{k=1}^{m+1} \frac{1}{2^{k^2}} & \leq 1 - \frac{1}{2^{m^2}} + \frac{1}{2^{(m+1)^2}} \\
& = 1 - \frac{1}{2^{(m+1)^2}} + \left(\frac{1}{2^{(m+1)^2}} - \frac{1}{2^{m^2}} + \frac{1}{2^{(m+1)^2}}\right) \\
& = 1 - \frac{1}{2^{(m+1)^2}} + \left(\frac{2}{2^{(m+1)^2}} - \frac{1}{2^{m^2}}\right) \\
& = 1 - \frac{1}{2^{(m+1)^2}} + \left(\frac{1}{2^{(m+1)^2 - 1}} - \frac{1}{2^{m^2}}\right) \\
& \lt 1 - \frac{1}{2^{(m+1)^2}}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Going to the last line can be done since $(m+1)^2 - 1 = m^2 + 2m \gt m^2$, so you have
$$\frac{1}{2^{(m+1)^2 - 1}} \lt \frac{1}{2^{m^2}} \implies \frac{1}{2^{(m+1)^2 - 1}} - \frac{1}{2^{m^2}} \lt 0 \tag{3}\label{eq3A}$$
This shows the problem you're trying to solve for is true for $n = m + 1$, thus completing the inductive process. You can now conclude that
$$\sum_{k=1}^{n} \frac{1}{2^{k^2}} \leq 1-\frac{1}{2^{n^2}} \tag{4}\label{eq4A}$$
is true for all integers $n \ge 1$.
