Determining (approximately) the amount of a strong base needed to change a buffer's pH by a certain number of units
You can calculate the quantity of sodium hydroxide, required to change the $\mathrm{pH}$ of your system by two unites,  by calculating the concentration of sodium ion in the charge balance equation:
$$\ce{[Na+] +[H+]=[OH- ] +[H2PO4^{-}]} \,\,\,\,(1)$$
In fact, you know the initial concentration of sodium ion in the buffer system before adding sodium hydroxide (from the known concentration of $\ce{NaH2PO4}$), and you calculate the final concentration of sodium ion  after adding sodium hydroxide by the above equation. So you can deduce the required quantity  of sodium hydroxide.
As for the  other ions in the charge balance equation, they are calculated as follows:
At first the  $\mathrm{pH}$ of the buffer system $\mathrm{pH=pK}_a=2.15$.

If you want to change the $\mathrm{pH}$ of your system by two unites, then the final $\mathrm{pH}$: $$\mathrm{pH}= 4.15\,\,\,\,(2)$$
Now you can estimate the concentrations of ion hydronium and ion hydroxide.

The use of Henderson-Hasselbalch equation gives: $$\ce{[H2PO4^- ]= 100[H3PO4 ]}\,\,\,\,(3)$$

The equation of matter conservation gives: $$\ce{[H2PO4^- ] + [H3PO4 ]= C_0=1 \mathrm{M}}$$ As the buffer is $25\,\mathrm{mL}$ of $1\,\mathrm{M}$ of each of the components, so there's a total of $1\,\mathrm{M}$ in the $50\,\mathrm{mL}$ solution due to dilution (not as you mentioned in your comment).
Now, you can calculate the concentration of ion $\ce{H2PO4^{-}}$
from equation (3) and (4):$$\ce{[H2PO4^- ] = \frac{100}{101}\mathrm{M}}$$


If we substitute the numerical values in equation (1):
$$\ce{[Na+] +10^{-4.15}=\frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{[Na+]_0 + [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$0.50 +\ce{ [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{ [Na+]_{\mathrm{added}}} \approx 0.4902 \mathrm{M} $$
$$n=\ce{  [Na+]_{\mathrm{added}}} \times V_{\mathrm{sol}}= 0.4902\times 50\times 10^{-3} =24.51 \times 10^{-3} \mathrm{mol} $$
The volume of sodium hydroxide is $$V_{\mathrm{added}} = 24.51 \times 10^{-3}\times 1= 24.51 \times 10^{-3} \mathrm{L} $$
I suggest to use a  solution of sodium hydroxide more concentrated, so as to neglect the effect of the added volume on the initial concentrations of the buffer system.